#include <bits/stdc++.h> // By Lucky Ox
#define int long long
#define endl "\n"
#define pii pair<int, int>
#define PI atan(1.0) * 4
using namespace std;
using i128 = __int128;
typedef unsigned long long ull;
const int mod = LONG_LONG_MAX, INF = 0x3f3f3f3f3f3f3f3f;
int P(int x, int p){ return (x % p + p) % p; }
int lcm(int x, int y) { return x / gcd(x, y) * y; }
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int len_10(int x) { int len = 0; while(x) { x /= 10; len ++ ; } return len; }
int q_pow(int a, int k, int p) { int res = 1; while (k) { if (k & 1) res = res * a % p; k >>= 1; a = a * a % p; } return res; }
int to_int(string s) { int val = 0; for(int i = 0; i < (int)s.size(); i ++ ){val *= 10; val += s[i] - '0';}return val;}//注意:s是空串也会返回0
i128 read() { i128 x = 0; char c = getchar(); while (c < '0' || c > '9') c = getchar(); while (c >= '0' && c <= '9') {
x = x * 10 + c - '0'; c = getchar(); } return x; } //i128输入
void print(i128 x) {if(x > 9) print(x / 10); putchar(x % 10 + '0'); }//i128输出
//用__lg()来求一个数二进制下的位数 返回的len 表示这个数是[0, 1, ...., len] 比如10 __lg(10) = 3, 1010 [3, 2, 1, 0]
//__builtin_popcountll(int x) 求二进制下x中1的数量 __buitlin_ctzll(int x) 求二进制下末尾0的个数
//(n & (1 << i))的值可能会是1, 2, 4...... (n >> i) & 1的值一定是1
//将x转换成[1, n]中对应的数 (x - 1) % n + 1
//bool __ED__;
const int N = 1e6 + 10;
int n;
string s;
//bool __ST__;
void solve() {
    cin >> s;

    if(count(s.begin(), s.end(), s[0]) == s.size()) {
        cout << s.size() / 2 << endl;
        return ;
    } 

    int pot = 0;
    for(int i = 0; i < s.size(); i ++ ) {
        if(s[i] == s.back()) {
            s.push_back(s[i]);
        }
        else {
            pot = i;
            break;
        }
    }
    s = s.substr(pot);
    //cout << s << endl;
    vector<int> a[2];
    int cnt = 0;
    for(int i = 0; i < s.size(); i ++ ) {
        if(i != 0 && s[i] != s[i - 1]) {
            a[cnt & 1].push_back(cnt);
            cnt = 0;
        }
        cnt ++ ;
    }
    if(cnt) a[cnt & 1].push_back(cnt);

    int ans = 0;
    for(int i = 0; i < a[1].size(); i ++ ) {
        ans += a[1][i] / 2;
    }

    for(int i = 0; i < a[0].size(); i ++ ) {
        ans += a[0][i] / 2;
    }
    ans -= (a[0].size() >= 1);
    cout << ans << endl;
}

signed main() {
    //cerr << abs(&__ED__-&__ST__) / (1024.0 * 1024.0) <<" MiB\n";
    ios::sync_with_stdio(0);cin.tie(0);
    cout << fixed << setprecision(10);
    int T = 1;
    cin >> T;
    while (T -- ) solve();
    return 0;
}