#include <bits/stdc++.h>
//#define int long long
#define endl "\n"
using namespace std;

using ll = long long;
const int INF = 1e9;
const int MOD = 1e9 + 7;
const int N = 5e5 + 5;

//  题意 :
//     给定一个数组，你可以选择任意的一个区间[l, r]
//  求  max{cnt[l, r] - mex[l, r]}

//  解法 :
//     枚举 mex [1-n], 可以划分出 n 个区间, 离线询问每个区间的 cnt[l, r], 树状数组实现

struct BIT {
    vector<ll> tr;
    int n;

    BIT(int n) : n(n) {
        tr.resize(n + 1, 0);
    }

    int lowbit(int x) {
        return x & -x;
    }

    void add(int x, int k) {
        for (int i = x; i <= n; i += lowbit(i)) {
            tr[i] += k;
        }
    }

    ll sum(int x) {
        ll res = 0;
        for (int i = x; i >= 1; i -= lowbit(i)) {
            res += tr[i];
        }
        return res;
    }
};

vector<tuple<int, int, int>> query;
vector<int> p[N];
int vis[N];
int a[N], b[N];

void solve() {
    int n, m;
    cin >> n >> m;
    query.clear();
    for (int i = 1; i <= n + 1; i++) {
        p[i].clear();
        p[i].emplace_back(0);
        vis[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        b[i] = a[i];
        if (a[i] > n)
            p[n + 1].emplace_back(i);
        else 
            p[a[i]].emplace_back(i);
    }
    
    for (int i = 1; i <= n + 1; i++) {
        p[i].emplace_back(n + 1);
        for (int j = 0; j < p[i].size() - 1; j++) {
            if (p[i][j] + 1 <= p[i][j + 1] - 1)
                query.emplace_back(p[i][j] + 1, p[i][j + 1] - 1, i);
        }
    }

    sort(b + 1, b + n + 1);
    for (int i = 1; i <= n; i++) {
        a[i] = lower_bound(b + 1, b + n + 1, a[i]) - b;
    }
    sort(query.begin(), query.end(), [&](const auto& q1, const auto& q2) {
        return get<1>(q1) < get<1>(q2);
    });
    ll ans = 0;
    BIT bit(2 * n);
    for (int i = 0, start = 1; i < query.size(); i++) {
        auto [l, r, mex] = query[i];
        for (int j = start; j <= r; j++) {
            if (vis[a[j]])
                bit.add(vis[a[j]], -1);
            bit.add(j, 1);
            vis[a[j]] = j;
        }
        start = r + 1;
        ans = max(ans, bit.sum(r) - bit.sum(l - 1) - mex);
    }
    cout << ans << endl;
}

int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);

    int T = 1;
    cin >> T;

    while (T--) {
        solve();
    }

    return 0;
}