#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void read(T& x) {
  int f = 0;
  x = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = f ? -x : x;
}
const int N = 1e5 + 10;
const int INF = 0x7fffffff;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1);
inline int tr(int x) { return (x - 1) >> 1; }
bool vst[N];
vector<int> a, b;
int main() {
  int t;
  read(t);
  while (t--) {
    memset(vst, 0, sizeof(vst));
    a.clear();
    b.clear();
    int n, cur, ans = 0;
    read(n);
    for (int i = 1; i <= n; i++) {
      read(cur);
      if (cur > n || vst[cur])
        a.push_back(cur);
      else
        vst[cur] = true;
    }
    for (int i = 1; i <= n; i++)
      if (!vst[i]) b.push_back(i);
    sort(a.begin(), a.end());
    cur = a.size();
    for (int i = 0; i < cur; i++, ans++)
      if (b[i] > a[i] || b[i] > tr(a[i])) {
        ans = -1;
        break;
      }
    printf("%d\n", ans);
  }
  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;
template <class T>
inline void read(T& x) {
  int f = 0;
  x = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = f ? -x : x;
}
const int N = 1e5 + 10;
const int INF = 0x7fffffff;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1);
inline int tr(int x) { return (x - 1) >> 1; }
bool vst[N];
vector<int> a, b;
int main() {
  int t;
  read(t);
  while (t--) {
    memset(vst, 0, sizeof(vst));
    a.clear();
    b.clear();
    int n, cur, ans = 0;
    read(n);
    for (int i = 1; i <= n; i++) {
      read(cur);
      if (cur > n || vst[cur])
        a.push_back(cur);
      else
        vst[cur] = true;
    }
    for (int i = 1; i <= n; i++)
      if (!vst[i]) b.push_back(i);
    sort(a.begin(), a.end());
    cur = a.size();
    for (int i = 0; i < cur; i++, ans++)
      if (b[i] > a[i] || b[i] > tr(a[i])) {
        ans = -1;
        break;
      }
    printf("%d\n", ans);
  }
  return 0;
}

```