/*
注意到从前往后考虑 $ S $ 串不容易维护。我们考虑从最后一个状态出发，最后的 $ S $ 串一定存在是一个字符的情况。然后我们构成出反串 $ S^R $ ，上一个状态可能是由 $ t S w S ^ R t ^ R $ 组成，这里有两种情况。

1. 所有情况，那么贪心考虑， $t$ 和 $t^R$ 一定不存在。我们只需要寻找最靠前的 $S^R$ 即可，左边界不变（$w$ 可能为空）。
2. 若 $ S $ 是回文串且 $ S $前面一个字符和后面一个字符相等，那么 $t S t$ 依旧是回文串，左边界左移1位。若$ t S $ 依旧是回文串，左边界左移1位。

注意到操作1的 $w$ 串为空时，新串为回文串，否则 $S$ 的长度至少翻倍。那么操作1在 $w$ 不为空的操作次数最多执行 $log$ 次。

注意到对于回文子串 $ T $ 可以写成 $ t w t^R $，在 $t$ 中选择的左端点进行操作1，一定能匹配到 $t^R$ 中的部分。若再执行操作2，一定劣于从 $w$ 开始执行操作2，且一定能执行出相同的左右端点。这里显然发现对于最多的操作次数一定为形如 $222...2111...1$ 。因为若操作1形成回文串，不如直接从回文中心开始构造。

再次注意到，对于回文串，考虑从回文串最左边开始匹配进行操作1，一定是匹配 $w$ 为空的新状态，即对于操作结束后相同的左端点可以执行操作2那么一定执行操作2。

那么我们用回文树/马拉车先维护出所有左端点开始操作最多的回文串。再暴力执行1操作，考虑次数即可。
*/
#pragma GCC optimize(2,"Ofast","inline")
#include <bits/stdc++.h>
#define ll long long
#define enl putchar('\n')
#define all(x) (x).begin(),(x).end()
#define debug(x) printf(" "#x":%d\n",x);
using namespace std;
const int MAXN = 2e5 + 5, LOG = 19;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
typedef pair<int, int> pii;
char buf[1 << 21], * p1 = buf, * p2 = buf, obuf[1 << 21], * o = obuf, of[35];
#define gc()(p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
inline ll qpow(ll a, ll n) { ll res = 1; while (n) { if (n & 1)res = res * a % mod; n >>= 1; a = a * a % mod; }return res; }
template <class T = int>inline T read() { T s = 0, f = 1; char c = gc(); for (; !isdigit(c); c = gc())if (c == '-')f = -1; for (; isdigit(c); c = gc())s = s * 10 + c - '0'; return s * f; }
inline void read(int* a, int n) { for (int i = 1; i <= n; ++i)a[i] = read(); }
inline int inal(char* s) { int n = 0; for (s[0] = gc(); !isalpha(s[0]); s[0] = gc()); for (; isalpha(s[n]); s[++n] = gc()); return s[n] = 0, n; }
inline int indi(char* s) { int n = 0; for (s[0] = gc(); !isdigit(s[0]); s[0] = gc()); for (; isdigit(s[n]); s[++n] = gc()); return s[n] = 0, n; }
inline void outd(auto* a, int n) { for (int i = 1; i <= n; ++i)printf("%d ", a[i]); enl; }
int n, m, q;
char s[MAXN], str[MAXN];
int f[MAXN], com[MAXN];

struct HJT {
    int T[MAXN], sum[MAXN * LOG], L[MAXN * LOG], R[MAXN * LOG];
    int tot, n;
    void clear(int n) {
        memset(T + 1, 0, sizeof(int) * n);
        // memset(sum + 1, 0, sizeof(int) * tot);
        // memset(L + 1, 0, sizeof(int) * tot);
        // memset(R + 1, 0, sizeof(int) * tot);
        tot = 0;
    }
    void insert(int l, int r, int p, int& u, const int& tar) {
        sum[++tot] = sum[p] + 1;
        L[tot] = L[p]; R[tot] = R[p];
        u = tot;
        if (l == r)return;
        int m = (l + r) >> 1;
        if (tar > m)insert(m + 1, r, R[p], R[u], tar);
        else insert(l, m, L[p], L[u], tar);
    }
    int query(const int& l, const int& r, const int& qv, const int& u, const int& p) {
        if (l == r)return l;
        int m = (l + r) >> 1, res = 0;
        int lv = sum[L[u]] - sum[L[p]], rv = sum[R[u]] - sum[R[p]];
        if (qv <= m && lv)res = query(l, m, qv, L[u], L[p]);
        if (!res && rv && r != n)res = query(m + 1, r, qv, R[u], R[p]);
        return res;
    }
};

struct Sa {     //俩字符串连一起记得开俩倍空间
    char s[MAXN];
    int SA[MAXN], rk[MAXN], oldrk[MAXN << 1], id[MAXN], key1[MAXN], cnt[MAXN];  //SA排名为i对应原字符串下标，rk下标为i的后缀排名
    int height[MAXN], n;    //height与上一个排名相同的个数，height[1]=0
    HJT hjt;
    void clear() {
        hjt.clear(n);
        memset(cnt, 0, sizeof(int) * max(n + 5, 128));
        memset(rk + 1, 0, sizeof(int) * n);
        // memset(cnt, 0, sizeof(cnt));
        // memset(SA, 0, sizeof(SA));
        // memset(oldrk, 0, sizeof(oldrk));
        // memset(id, 0, sizeof(id));
        // memset(key1, 0, sizeof(key1));
        // memset(height, 0, sizeof(height));
        // memset(rk, 0, sizeof(rk));
    }
    inline bool cmp(int x, int y, int w) {
        return oldrk[x] == oldrk[y] && oldrk[x + w] == oldrk[y + w];
    }
    inline void getSA() {
        int m = 127;
        for (int i = 1; i <= n; ++i)++cnt[rk[i] = s[i]];
        for (int i = 1; i <= m; ++i)cnt[i] += cnt[i - 1];
        for (int i = n; i >= 1; --i)SA[cnt[rk[i]]--] = i;
        for (int len = 1, p;; len <<= 1, m = p) {
            p = 0;
            for (int i = n; i > n - len; --i)id[++p] = i;
            for (int i = 1; i <= n; ++i)
                if (SA[i] > len)id[++p] = SA[i] - len;
            // memset(cnt, 0, sizeof(cnt));
            memset(cnt + 1, 0, sizeof(int) * m);
            // memset(cnt + 1, 0, sizeof(int) * max(n, 127));
            for (int i = 1; i <= n; ++i)++cnt[key1[i] = rk[id[i]]];
            for (int i = 1; i <= m; ++i)cnt[i] += cnt[i - 1];
            for (int i = n; i >= 1; --i)SA[cnt[key1[i]]--] = id[i];
            memcpy(oldrk + 1, rk + 1, n * sizeof(int));
            p = 0;
            for (int i = 1; i <= n; ++i)
                rk[SA[i]] = cmp(SA[i], SA[i - 1], len) ? p : ++p;
            if (p == n)break;
        }
        hjt.n = n;
        for (int i = 1; i <= n; ++i)
            hjt.insert(1, n, hjt.T[i - 1], hjt.T[i], SA[i]);
        // memset(cnt + 1, 0, sizeof(int) * n);
        // for (int i = 1; i <= n; ++i)printf("%d ", SA[i]); printf("\n");
        // for (int i = 1; i <= n; ++i)printf("%d ", rk[i]); printf("\n");
    }

    inline void getHeight() {
        for (int i = 1, k = 0; i <= n; ++i) {
            if (rk[i] == 0)continue;
            if (k)--k;
            while (s[i + k] == s[SA[rk[i] - 1] + k])++k;
            height[rk[i]] = k;
        }
    }
    int st[MAXN][LOG], lg[MAXN];
    inline void initST() {
        lg[1] = 0;
        for (int i = 2; i <= n; i++)lg[i] = lg[i >> 1] + 1;
        for (int i = 1; i <= n; i++)st[i][0] = height[i];
        for (int j = 1; j <= lg[n]; j++)
            for (int i = 1; i <= n - (1 << j) + 1; i++)
                st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
    }
    inline int RMQ(int x, int y) {
        if (x > y)return 0;
        int t = lg[y - x + 1];
        return min(st[x][t], st[y - (1 << t) + 1][t]);
    }
    inline int LCP(int x, int y) {
        x = rk[x], y = rk[y];
        if (x > y)swap(x, y);
        return RMQ(x + 1, y);
    }
    int findL(int p, int len) {
        int l = 1, r = p;
        while (l < r) {
            int mid = l + r >> 1;
            if (RMQ(mid + 1, p) >= len)r = mid;
            else l = mid + 1;
        }
        return r;
    }
    int findR(int p, int len) {
        int l = p, r = n;
        while (l < r) {
            int mid = l + r + 1 >> 1;
            if (RMQ(p + 1, mid) >= len)l = mid;
            else r = mid - 1;
        }
        return l;
    }
    int find_nxt(int l, int r) {
        int len = r - l + 1;
        int L = findL(rk[n - r + 1], len), R = findR(rk[n - r + 1], len);
        int res = hjt.query(1, n, r + 1, hjt.T[L - 1], hjt.T[R]);
        return (res && res <= n / 2) ? res + len - 1 : 0;
    }
}p;



void manacher(char* str, int n) {
    int t = 0;
    for (int i = 1; i <= n; ++i) {
        if (t + f[t] >= i)f[i] = min(f[t - (i - t)], t + f[t] - i);
        while (i - f[i] - 1 > 0 && i + f[i] + 1 <= n && str[i - f[i] - 1] == str[i + f[i] + 1])f[i]++;
        if (f[i] + i > f[t] + t) t = i;
    }
}
vector<array<int, 3>> op[MAXN];

void solve() {
    n = inal(s + 1);
    // printf("%s\n", s + 1);
    // for (int i = 1; i <= n; ++i)printf("%d", i % 10); enl;
    memcpy(p.s + 1, s + 1, sizeof(char) * n);
    p.s[n + 1] = '#';
    for (int i = 1; i <= n; ++i) p.s[n + i + 1] = s[n - i + 1];
    p.n = n + n + 1;
    p.clear();
    p.getSA();
    p.getHeight();
    p.initST();

    int ans = 0;
    for (int i = 1; i <= n; ++i) op[i].clear();
    for (int i = n; i >= 1; --i)
        com[i] = s[i] == s[i + 1] ? com[i + 1] + 1 : 0;
    str[m = 1] = '#';
    for (int i = 1; i <= n; ++i)
        str[++m] = s[i], str[++m] = '#';
    memset(f + 1, 0, sizeof(int) * m);
    manacher(str, m);
    // outd(f, m);
    for (int i = 1; i <= m; ++i) {
        int l = (i - f[i] + 1) / 2, c = (i + 1) / 2, r = (i + f[i]) / 2;
        int f = i & 1 ? -1 : 1;
        int v = min(c + com[c], r);
        op[l].push_back({v, f * c, 1});
        op[c].push_back({v, f * c, 0});
    }
    set<pii>S;
    for (int i = 1; i <= n; ++i) {
        for (auto x : op[i]) {
            if (x[2])S.insert({x[0], x[1]});
            else S.erase(S.find({x[0], x[1]}));
        }
        int v = 0, c = i;
        if (S.size()) {
            auto it = *S.rbegin();
            v = it.first - i;
            if (it.second > 0) c = it.second + it.second - i;
            else c = -it.second - it.second - i - 1;
            c = max(c, it.first);
        }
        while (c = p.find_nxt(i, c))
            ++v;
        ans = max(ans, v);
    }
    printf("%d\n", ans);
    memset(p.s + 1, 0, sizeof(char) * m);
}
signed main(signed argc, char const* argv[]) {
    clock_t c1 = clock();
#ifdef LOCAL
    freopen("in.in", "r", stdin);
    freopen("out.out", "w", stdout);
#endif
    //=============================================================
    int TxT = 1;
    TxT = read();
    while (TxT--)
        solve();
    //=============================================================
#ifdef LOCAL
    end :
    cerr << "Time Used:" << clock() - c1 << "ms" << endl;
#endif
    return 0;
}