#include <bits/stdc++.h>
using namespace std;
inline void intxt() {}
inline void read(int &x) {
  int f = 1;
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
inline void read(long long &x) {
  long long f = 1;
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
const long long mod = 998244353;
const int N = 4e5 + 5;
const int M = 1e6 + 5;
int n, m, k;
int a[N];
multiset<int> cup;
int main() {
  intxt();
  int casen;
  scanf("%d", &casen);
  for (int LP = 1; LP <= casen; ++LP) {
    cin >> n;
    cup.clear();
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      cup.insert(a[i]);
    }
    int ans = 0;
    for (int i = n; i >= 1; i--) {
      if (cup.count(i) >= 1) {
        auto x = cup.lower_bound(i);
        cup.erase(x);
      } else {
        auto x = cup.upper_bound(i * 2);
        if (x == cup.end()) {
          ans = -1;
          break;
        }
        cup.erase(x);
        ans++;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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>
using namespace std;
inline void intxt() {}
inline void read(int &x) {
  int f = 1;
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
inline void read(long long &x) {
  long long f = 1;
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
const long long mod = 998244353;
const int N = 4e5 + 5;
const int M = 1e6 + 5;
int n, m, k;
int a[N];
multiset<int> cup;
int main() {
  intxt();
  int casen;
  scanf("%d", &casen);
  for (int LP = 1; LP <= casen; ++LP) {
    cin >> n;
    cup.clear();
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      cup.insert(a[i]);
    }
    int ans = 0;
    for (int i = n; i >= 1; i--) {
      if (cup.count(i) >= 1) {
        auto x = cup.lower_bound(i);
        cup.erase(x);
      } else {
        auto x = cup.upper_bound(i * 2);
        if (x == cup.end()) {
          ans = -1;
          break;
        }
        cup.erase(x);
        ans++;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```