description: 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.
solution: #include <bits/stdc++.h>
  int cmp(const void *p, const void *q) { return (*(int *)p) - (*(int *)q); }
  int a[100010];
  int need[100010];
  int main() {
    int t, n;
    int i;
    int m;
    int ans;
    scanf("%d", &t);
    while (t--) {
      scanf("%d", &n);
      ans = 0;
      m = n;
      for (i = 1; i <= n; i++) need[i] = 0;
      for (i = 1; i <= n; i++) scanf("%d", &a[i]);
      for (i = 1; i <= n; i++) {
        if (a[i] <= n) {
          if (need[a[i]] == 0) {
            need[a[i]] = 1;
            a[i] = 0;
          }
        }
      }
      qsort(a + 1, n, sizeof(a[0]), cmp);
      for (i = 1; a[i] == 0 && i <= n; i++)
        ;
      m = i;
      for (i = 1; i <= n; i++) {
        if (need[i] == 1) continue;
        if (a[m] <= 2 * i) {
          printf("-1\n");
          ans = -1;
          break;
        }
        ans++;
        m++;
      }
      if (ans >= 0) printf("%d\n", ans);
    }
  }
  