#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 100;
const int inf = 1e9;
const long long mod = 998244353;
int n;
int a[N];
int hav[N];
void sol() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) hav[i] = 0;
  vector<int> can;
  int res = n;
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    if (x >= 1 && x <= n) {
      if (!hav[x]) {
        hav[x] = 1;
        res--;
      } else
        can.push_back((x - 1) / 2);
    } else
      can.push_back((x - 1) / 2);
  }
  sort((can).begin(), (can).end());
  int nw = n;
  while (nw >= 1) {
    while (nw >= 1 && hav[nw]) --nw;
    if (nw == 0) break;
    if (can.back() < nw) return (void)(cout << -1 << '\n');
    can.pop_back();
    hav[nw] = 1;
  }
  cout << res << '\n';
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) sol();
}

### Prompt

Please provide a CPP coded solution to the problem described below:

Paprika loves permutations. She has an array a_1, a_2, ..., a_n. She wants to make the array a permutation of integers 1 to n.

In order to achieve this goal, she can perform operations on the array. In each operation she can choose two integers i (1 ≤ i ≤ n) and x (x > 0), then perform a_i := a_i mod x (that is, replace a_i by the remainder of a_i divided by x). In different operations, the chosen i and x can be different.

Determine the minimum number of operations needed to make the array a permutation of integers 1 to n. If it is impossible, output -1.

A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

Input

Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.

The first line of each test case contains an integer n (1 ≤ n ≤ 10^5).

The second line of each test case contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^9).

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output the minimum number of operations needed to make the array a permutation of integers 1 to n, or -1 if it is impossible.

Example

Input


4
2
1 7
3
1 5 4
4
12345678 87654321 20211218 23571113
9
1 2 3 4 18 19 5 6 7


Output


1
-1
4
2

Note

For the first test, the only possible sequence of operations which minimizes the number of operations is: 

  * Choose i=2, x=5. Perform a_2 := a_2 mod 5 = 2. 



For the second test, it is impossible to obtain a permutation of integers from 1 to n.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 100;
const int inf = 1e9;
const long long mod = 998244353;
int n;
int a[N];
int hav[N];
void sol() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) hav[i] = 0;
  vector<int> can;
  int res = n;
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    if (x >= 1 && x <= n) {
      if (!hav[x]) {
        hav[x] = 1;
        res--;
      } else
        can.push_back((x - 1) / 2);
    } else
      can.push_back((x - 1) / 2);
  }
  sort((can).begin(), (can).end());
  int nw = n;
  while (nw >= 1) {
    while (nw >= 1 && hav[nw]) --nw;
    if (nw == 0) break;
    if (can.back() < nw) return (void)(cout << -1 << '\n');
    can.pop_back();
    hav[nw] = 1;
  }
  cout << res << '\n';
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) sol();
}

```