#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 1e18;
int nx[4] = {1, -1, 0, 0}, ny[4] = {0, 0, 1, -1};
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
void solve() {
  long long n;
  cin >> n;
  vector<long long> A;
  A.assign(n, 0);
  set<long long> s;
  bool vis[n];
  memset(vis, false, sizeof(vis));
  for (int i = 0; i < n; i++) {
    s.insert(i + 1);
    cin >> A[i];
  }
  sort((A).begin(), (A).end());
  for (int i = 0; i < n; i++) {
    if (s.count(A[i]) != 0) {
      s.erase(A[i]);
      vis[i] = true;
    }
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (vis[i]) continue;
    long long main = *(s.begin());
    if (A[i] != main && A[i] / 2 + (A[i] & 1) <= main) {
      cout << -1 << '\n';
      return;
    } else if (A[i] != main) {
      cnt++;
    }
    s.erase(s.begin());
  }
  cout << cnt << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tt;
  cin >> tt;
  while (tt--) solve();
}

### Prompt

Generate 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>
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 1e18;
int nx[4] = {1, -1, 0, 0}, ny[4] = {0, 0, 1, -1};
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
void solve() {
  long long n;
  cin >> n;
  vector<long long> A;
  A.assign(n, 0);
  set<long long> s;
  bool vis[n];
  memset(vis, false, sizeof(vis));
  for (int i = 0; i < n; i++) {
    s.insert(i + 1);
    cin >> A[i];
  }
  sort((A).begin(), (A).end());
  for (int i = 0; i < n; i++) {
    if (s.count(A[i]) != 0) {
      s.erase(A[i]);
      vis[i] = true;
    }
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (vis[i]) continue;
    long long main = *(s.begin());
    if (A[i] != main && A[i] / 2 + (A[i] & 1) <= main) {
      cout << -1 << '\n';
      return;
    } else if (A[i] != main) {
      cnt++;
    }
    s.erase(s.begin());
  }
  cout << cnt << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tt;
  cin >> tt;
  while (tt--) solve();
}

```