https://codeforces.com/contest/1972/problem/C

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PII;
typedef vector<long long> VI;
#define int long long
#define INF 0x3f3f3f3f
#define endl '\n'
#define N 200010
const int mod = 1e9 + 7;

ll ksm(ll a, ll b) {
    ll ans = 1;
    for (; b; b >>= 1LL) {
        if (b & 1) ans = ans * a % mod;
        a = a * a % mod;
    }
    return ans;
}
ll lowbit(ll x) {
    return -x & x;
}

int p[N], si[N];
int find(int x) {
    if (x == p[x])return p[x];
    p[x] = find(p[x]);
    return p[x];
}
//size[find(b)] += size[find(a)];
//p[find(a)] = find(b);
int n, k;
vector<int> a(N);

int check(int x) { //二分判断k能否满足分配后ai均大于等于x
    int need = 0;

    for (int i = 1; i <= n; i ++) {
        if (a[i] < x)need += x - a[i];
    }

    return need <= k;
}

void solve() {

    cin >> n >> k;


    for (int i = 1; i <= n; i ++)cin >> a[i];

    int l = 0, r = 2e12;
    int mx = 0;

    while (l + 1 < r) {  // 二分找最大的最低分配后的值
        int mid = (l + r) / 2;
        if (check(mid))l = mid;
        else r = mid;
    }

    if (check(l))mx = l;
    else mx = r;

    for (int i = 1; i <= n; i ++) {
        if (a[i] < mx) {
            int t = mx - a[i];
            a[i] += t;
            k -= t;
        }
    }

    int num = 0;

    for (int i = 1; i <= n ; i ++) {
        if (a[i] == mx && k) {
            a[i] ++;
            k --;
        }

        if (a[i] > mx)num ++;
    }
    // 1 2 3 4 1 2 3 4 1 2 3 4  
    int res = 1 + (mx - 1) * n + num; //贪心算排列数 1 + (mx - 1) * n + 多出来数的种类数
    cout << res << endl;
}

signed main() {

    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    cin >> T;
    while (T --)

        solve();

    return 0;
}