#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const int N = 1e5 + 5;
const long long int mod = 1e9 + 7;
const long long int inf = 1e17;
const long double PI = (acos(-1));
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> arr(n);
    map<int, int> mp;
    for (int i = 0; i < n; ++i) {
      cin >> arr[i];
      ++mp[arr[i]];
    }
    bool flag = true;
    vector<int> res(n, 0);
    for (int i = 0; i < n; ++i) {
      if (mp.find(i + 1) != mp.end()) {
        --mp[i + 1];
        res[i] = i + 1;
      }
    }
    auto it = mp.begin();
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      if (res[i] > 0) {
        continue;
      }
      while (it != mp.end() && it->second <= 0) {
        ++it;
      }
      if (it != mp.end() && (2 * (i + 1) < (it->first))) {
        ++ans;
        res[i] = i + 1;
        --(it->second);
      }
      if (res[i] == 0) {
        flag = false;
        break;
      }
    }
    if (!flag) {
      ans = -1;
    }
    cout << ans << endl;
  }
}

### 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const int N = 1e5 + 5;
const long long int mod = 1e9 + 7;
const long long int inf = 1e17;
const long double PI = (acos(-1));
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> arr(n);
    map<int, int> mp;
    for (int i = 0; i < n; ++i) {
      cin >> arr[i];
      ++mp[arr[i]];
    }
    bool flag = true;
    vector<int> res(n, 0);
    for (int i = 0; i < n; ++i) {
      if (mp.find(i + 1) != mp.end()) {
        --mp[i + 1];
        res[i] = i + 1;
      }
    }
    auto it = mp.begin();
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      if (res[i] > 0) {
        continue;
      }
      while (it != mp.end() && it->second <= 0) {
        ++it;
      }
      if (it != mp.end() && (2 * (i + 1) < (it->first))) {
        ++ans;
        res[i] = i + 1;
        --(it->second);
      }
      if (res[i] == 0) {
        flag = false;
        break;
      }
    }
    if (!flag) {
      ans = -1;
    }
    cout << ans << endl;
  }
}

```