#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
long long powe(T base, T to) {
  long long ans = 1;
  while (to > 0) {
    if (to & 1)
      ans = ((ans) * (base)), to--;
    else
      base = ((base) * (base)), to /= 2;
  }
  return ans;
}
template <typename T>
T gcd(T a, T b) {
  return (b == 0) ? a : gcd(b, a % b);
}
bool prime(long long n) {
  if (n == 1) return 0;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return 0;
  return 1;
}
int binarySearch(int arr[], int start, int end, int val) {
  int mid;
  end--;
  while (start <= end) {
    mid = start + (end - start) / 2;
    if (arr[mid] == val)
      return mid;
    else if (arr[mid] < val)
      start = mid + 1;
    else
      end = mid - 1;
  }
  return -1;
}
long long fun(int arr[], int n, int x) {
  long long ans = x;
  for (int i = 0; i < n - 1; i++) ans += min(x, arr[i + 1] - arr[i]);
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tl = 1;
  cin >> tl;
  while (tl--) {
    int n;
    cin >> n;
    int arr[n];
    for (auto &i : arr) cin >> i;
    ;
    sort(arr, arr + n);
    bool ok = true;
    for (int i = 0; i < n; i++)
      if (arr[i] < (i + 1)) {
        ok = false;
        break;
      }
    if (!ok) {
      cout << "-1\n";
      continue;
    }
    vector<bool> tmp(n, 0);
    int ans = n;
    int val = 1;
    for (int i = 0; i < n; i++) {
      if (arr[i] >= 1 && arr[i] <= n && !tmp[arr[i] - 1]) {
        tmp[arr[i] - 1] = 1, ans--;
        while (tmp[val - 1] && val <= n) val++;
        continue;
      }
      if (arr[i] == val) {
        tmp[val - 1] = 1;
        while (tmp[val - 1] && val <= n) val++;
        continue;
      }
      if ((arr[i] - val) <= val) {
        ans = -1;
        break;
      }
      tmp[val - 1] = 1;
      while (tmp[val - 1] && val <= n) val++;
    }
    cout << ans << "\n";
  }
}

### Prompt

Please formulate a CPP solution to 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>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
long long powe(T base, T to) {
  long long ans = 1;
  while (to > 0) {
    if (to & 1)
      ans = ((ans) * (base)), to--;
    else
      base = ((base) * (base)), to /= 2;
  }
  return ans;
}
template <typename T>
T gcd(T a, T b) {
  return (b == 0) ? a : gcd(b, a % b);
}
bool prime(long long n) {
  if (n == 1) return 0;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return 0;
  return 1;
}
int binarySearch(int arr[], int start, int end, int val) {
  int mid;
  end--;
  while (start <= end) {
    mid = start + (end - start) / 2;
    if (arr[mid] == val)
      return mid;
    else if (arr[mid] < val)
      start = mid + 1;
    else
      end = mid - 1;
  }
  return -1;
}
long long fun(int arr[], int n, int x) {
  long long ans = x;
  for (int i = 0; i < n - 1; i++) ans += min(x, arr[i + 1] - arr[i]);
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tl = 1;
  cin >> tl;
  while (tl--) {
    int n;
    cin >> n;
    int arr[n];
    for (auto &i : arr) cin >> i;
    ;
    sort(arr, arr + n);
    bool ok = true;
    for (int i = 0; i < n; i++)
      if (arr[i] < (i + 1)) {
        ok = false;
        break;
      }
    if (!ok) {
      cout << "-1\n";
      continue;
    }
    vector<bool> tmp(n, 0);
    int ans = n;
    int val = 1;
    for (int i = 0; i < n; i++) {
      if (arr[i] >= 1 && arr[i] <= n && !tmp[arr[i] - 1]) {
        tmp[arr[i] - 1] = 1, ans--;
        while (tmp[val - 1] && val <= n) val++;
        continue;
      }
      if (arr[i] == val) {
        tmp[val - 1] = 1;
        while (tmp[val - 1] && val <= n) val++;
        continue;
      }
      if ((arr[i] - val) <= val) {
        ans = -1;
        break;
      }
      tmp[val - 1] = 1;
      while (tmp[val - 1] && val <= n) val++;
    }
    cout << ans << "\n";
  }
}

```