#include <bits/stdc++.h>
#pragma GCC diagnostic ignored "-Wunused-variable"
#pragma GCC diagnostic ignored "-Wunused-parameter"
using namespace std;
#pragma GCC diagnostic ignored "-Wunused-but-set-variable"
#pragma GCC diagnostic ignored "-Wformat"
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long power(long long B, long long P) {
  if (P == 0) return 1;
  long long X = power(B, P / 2);
  if (P % 2 == 0)
    return X * X;
  else
    return B * X * X;
}
int fx[] = {1, -1, 0, 0};
int fy[] = {0, 0, 1, -1};
int main() {
  int T, cs = 0;
  scanf("%d", &T);
  while (T--) {
    int n;
    scanf("%d", &n);
    set<int> st;
    vector<int> v;
    for (int i = 1; i <= n; i++) st.insert(i);
    for (int i = 1; i <= n; i++) {
      int u;
      scanf("%d", &u);
      if (st.find(u) != st.end())
        st.erase(u);
      else
        v.push_back(u);
    }
    sort(v.rbegin(), v.rend());
    int res = 0;
    for (auto it : v) {
      auto ck = st.end();
      ck--;
      if (*ck > (it - 1) / 2) {
        res = 1;
        break;
      }
      st.erase(ck);
    }
    if (res == 1)
      cout << "-1\n";
    else
      cout << v.size() << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 diagnostic ignored "-Wunused-variable"
#pragma GCC diagnostic ignored "-Wunused-parameter"
using namespace std;
#pragma GCC diagnostic ignored "-Wunused-but-set-variable"
#pragma GCC diagnostic ignored "-Wformat"
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long power(long long B, long long P) {
  if (P == 0) return 1;
  long long X = power(B, P / 2);
  if (P % 2 == 0)
    return X * X;
  else
    return B * X * X;
}
int fx[] = {1, -1, 0, 0};
int fy[] = {0, 0, 1, -1};
int main() {
  int T, cs = 0;
  scanf("%d", &T);
  while (T--) {
    int n;
    scanf("%d", &n);
    set<int> st;
    vector<int> v;
    for (int i = 1; i <= n; i++) st.insert(i);
    for (int i = 1; i <= n; i++) {
      int u;
      scanf("%d", &u);
      if (st.find(u) != st.end())
        st.erase(u);
      else
        v.push_back(u);
    }
    sort(v.rbegin(), v.rend());
    int res = 0;
    for (auto it : v) {
      auto ck = st.end();
      ck--;
      if (*ck > (it - 1) / 2) {
        res = 1;
        break;
      }
      st.erase(ck);
    }
    if (res == 1)
      cout << "-1\n";
    else
      cout << v.size() << endl;
  }
  return 0;
}

```