#include <bits/stdc++.h>
using namespace std;
vector<int> v1[200005];
bool visit[200005];
bool color[200005];
int parent[200005];
int zz, zz1;
vector<int> vv;
void solve() {
  long long int n, m;
  cin >> n;
  vector<long long int> v1(n), store;
  set<long long int> set1;
  for (long long int x = 1; x < n + 1; ++x) set1.insert(x);
  for (long long int x = 0; x < n; ++x) {
    cin >> m;
    if (set1.count(m) > 0)
      set1.erase(m);
    else
      store.push_back(m);
  }
  sort(store.begin(), store.end());
  long long int count = 0, f1 = 1;
  for (long long int x = 0; x < store.size(); ++x) {
    long long int a = store[x];
    auto it = set1.begin();
    long long int pehla_ele = *it;
    long long int garaj = (a - 1) / 2;
    if (pehla_ele > garaj) {
      cout << -1 << endl;
      return;
    } else {
      set1.erase(pehla_ele);
      count++;
    }
  }
  cout << count << endl;
}
void test_case() {
  long long int t;
  cin >> t;
  while (t-- > 0) {
    solve();
  }
}
int main() { test_case(); }

### Prompt

Your challenge is to write 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;
vector<int> v1[200005];
bool visit[200005];
bool color[200005];
int parent[200005];
int zz, zz1;
vector<int> vv;
void solve() {
  long long int n, m;
  cin >> n;
  vector<long long int> v1(n), store;
  set<long long int> set1;
  for (long long int x = 1; x < n + 1; ++x) set1.insert(x);
  for (long long int x = 0; x < n; ++x) {
    cin >> m;
    if (set1.count(m) > 0)
      set1.erase(m);
    else
      store.push_back(m);
  }
  sort(store.begin(), store.end());
  long long int count = 0, f1 = 1;
  for (long long int x = 0; x < store.size(); ++x) {
    long long int a = store[x];
    auto it = set1.begin();
    long long int pehla_ele = *it;
    long long int garaj = (a - 1) / 2;
    if (pehla_ele > garaj) {
      cout << -1 << endl;
      return;
    } else {
      set1.erase(pehla_ele);
      count++;
    }
  }
  cout << count << endl;
}
void test_case() {
  long long int t;
  cin >> t;
  while (t-- > 0) {
    solve();
  }
}
int main() { test_case(); }

```