#include <bits/stdc++.h>
using namespace std;
long long int pc;
void PrintCase() { printf("Case #%lli: ", ++pc); }
long long int bigmod(long long int n, long long int p) {
  if (p == 0) return 1;
  if (p == 1) return (n + 1000000007) % 1000000007;
  if (p % 2)
    return (bigmod(n, p - 1) * n + 1000000007) % 1000000007;
  else {
    long long int x = bigmod(n, p / 2);
    return (x * x + 1000000007) % 1000000007;
  }
}
long long int power(long long int n, long long int p) {
  if (p == 0) return 1;
  if (p == 1) return n;
  if (p % 2)
    return power(n, p - 1) * n;
  else {
    long long int x = power(n, p / 2);
    return x * x;
  }
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, i, j, k;
    cin >> n;
    long long int a[n + 5], vis[n + 5];
    memset(vis, 0, sizeof(vis));
    vector<long long int> v;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      if (a[i] <= n and vis[a[i]] == 0) {
        vis[a[i]] = 1;
      } else
        v.push_back(a[i]);
    }
    sort(v.begin(), v.end());
    long long int ok = 0;
    for (i = 1, j = 0; i <= n and j < v.size(); i++) {
      if (vis[i]) continue;
      long long int x = v[j];
      if (x % (x - i) == i) {
        vis[i] = 1;
        ok++;
      } else {
        ok = -1;
        break;
      }
      j++;
    }
    cout << ok << endl;
  }
}

### Prompt

Create a solution in cpp for the following problem:

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;
long long int pc;
void PrintCase() { printf("Case #%lli: ", ++pc); }
long long int bigmod(long long int n, long long int p) {
  if (p == 0) return 1;
  if (p == 1) return (n + 1000000007) % 1000000007;
  if (p % 2)
    return (bigmod(n, p - 1) * n + 1000000007) % 1000000007;
  else {
    long long int x = bigmod(n, p / 2);
    return (x * x + 1000000007) % 1000000007;
  }
}
long long int power(long long int n, long long int p) {
  if (p == 0) return 1;
  if (p == 1) return n;
  if (p % 2)
    return power(n, p - 1) * n;
  else {
    long long int x = power(n, p / 2);
    return x * x;
  }
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, i, j, k;
    cin >> n;
    long long int a[n + 5], vis[n + 5];
    memset(vis, 0, sizeof(vis));
    vector<long long int> v;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      if (a[i] <= n and vis[a[i]] == 0) {
        vis[a[i]] = 1;
      } else
        v.push_back(a[i]);
    }
    sort(v.begin(), v.end());
    long long int ok = 0;
    for (i = 1, j = 0; i <= n and j < v.size(); i++) {
      if (vis[i]) continue;
      long long int x = v[j];
      if (x % (x - i) == i) {
        vis[i] = 1;
        ok++;
      } else {
        ok = -1;
        break;
      }
      j++;
    }
    cout << ok << endl;
  }
}

```