#include <bits/stdc++.h>
using namespace std;
#define fast_io ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define int long long
#define pb push_back
#define ff first
#define ss second
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((int)(x).size())
#define endl '\n'
#define yes cout << "yes\n"
#define no cout << "no\n"
#define rep(i,a,b) for(int i=a;i<b;++i)
#define per(i,a,b) for(int i=b-1;i>=a;--i)
#define each(x, a) for (auto& x : a)
const int INF = 1e18;
const int MOD = 1e9+7;
const int N = 2e5 + 5;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) { return (a / gcd(a, b)) * b; }
int power(int a, int b, int m = MOD) {
int res = 1;
while (b > 0) {
if (b & 1) res = res * a % m;
a = a * a % m;
b >>= 1;
}
return res;
}
int modinv(int a, int m = MOD) {
return power(a, m - 2, m);
}
// Check function: can we achieve the multiset 'b' using segments of max length 'x',
// with a total of 'n' operations?
bool check(int x, int n, const vector<int>& positive_b) {
if (x == 0) return false;
int total_required_ops = 0;
// Sum of ceil(b_i / x) for all positive elements b_i
each(val, positive_b) {
// ceil(val / x) = (val + x - 1) / x
total_required_ops += (val + x - 1) / x;
}
// The number of required operations (R) must be less than or equal to the total available operations (n).
return total_required_ops <= n;
}
void solve() {
int n;
cin >> n;
vector<int> b(n);
vector<int> positive_b;
rep(i, 0, n) {
cin >> b[i];
if (b[i] > 0) {
positive_b.pb(b[i]);
}
}
// Binary search for the maximum possible length x in [1, n].
int low = 1;
int high = n;
int ans = 0;
while (low <= high) {
int mid = low + (high - low) / 2;
if (check(mid, n, positive_b)) {
ans = mid;
low = mid + 1; // Try for a larger length
} else {
high = mid - 1; // Must use a smaller length
}
}
cout << ans << endl;
}
int32_t main() {
fast_io;
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
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