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.
8
2
2 3
3
3 4 5
1
1
1
2
1
3
1
4
1
12071207
5
5 5 5 9 14

1
2
0
-1
1
1
1
4

#include <bits/stdc++.h>
using namespace std;
const unsigned int M = 1000000007;
long long binpow(long long a, long long b);
long long modmul(long long a, long long b, long long m);
void solve() {
  long long n, x, ans = 0;
  cin >> n;
  multiset<long long> st;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    st.insert(x);
  }
  for (x = n; x > 0; x--) {
    auto it = st.find(x);
    if (it != st.end()) {
      st.erase(it);
      continue;
    }
    it--;
    if (*it < 2 * x + 1) {
      cout << "-1\n";
      return;
    }
    st.erase(it);
    ans++;
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % M;
    a = a * a % M;
    b >>= 1;
  }
  return res;
}
long long modmul(long long a, long long b, long long m) {
  return ((a % m) * (b % m)) % m;
}
