#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
inline void Prin(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) Prin(x / 10);
  putchar(x % 10 + '0');
}
const int qs = 1e5 + 7;
int T, n;
int a[qs], u[qs];
int b[qs];
int main() {
  T = read();
  while (T--) {
    n = read();
    int cnt = 0, flag = 1;
    for (int i = 1; i <= n; ++i) u[i] = 0;
    for (int i = 1; i <= n; ++i) a[i] = read();
    for (int i = 1; i <= n; ++i) {
      if (a[i] >= 1 && a[i] <= n && !u[a[i]]) {
        u[a[i]] = 1;
      } else
        b[++cnt] = a[i];
    }
    sort(b + 1, b + 1 + cnt);
    int j = 1;
    for (int i = 1; i <= n; ++i) {
      if (!u[i]) {
        int p = (b[j] + 1) / 2;
        j++;
        if (i >= p) {
          flag = 0;
        }
      }
    }
    if (!flag)
      cout << "-1\n";
    else
      cout << cnt << "\n";
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
inline void Prin(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) Prin(x / 10);
  putchar(x % 10 + '0');
}
const int qs = 1e5 + 7;
int T, n;
int a[qs], u[qs];
int b[qs];
int main() {
  T = read();
  while (T--) {
    n = read();
    int cnt = 0, flag = 1;
    for (int i = 1; i <= n; ++i) u[i] = 0;
    for (int i = 1; i <= n; ++i) a[i] = read();
    for (int i = 1; i <= n; ++i) {
      if (a[i] >= 1 && a[i] <= n && !u[a[i]]) {
        u[a[i]] = 1;
      } else
        b[++cnt] = a[i];
    }
    sort(b + 1, b + 1 + cnt);
    int j = 1;
    for (int i = 1; i <= n; ++i) {
      if (!u[i]) {
        int p = (b[j] + 1) / 2;
        j++;
        if (i >= p) {
          flag = 0;
        }
      }
    }
    if (!flag)
      cout << "-1\n";
    else
      cout << cnt << "\n";
  }
  return 0;
}

```