对于恰好就在 $[1, n]$ 范围内的数字，可以不用操作。因为一次的模数 $x$ 是任意取的，考虑 $a_i \bmod x$ 的取值范围，当 $a_i < x$ 的时候，$a_i \bmod x = a_i$；当 $a_i > x$ 时，$a_i \bmod x \le \lfloor \frac{a_i - 1}{2} \rfloor$，同时也能发现此时 $a_i \bmod x$ 的值域恰好是 $[1, \lfloor \frac{a_i - 1}{2} \rfloor]$ 中的所有整数。所以把剩余的 $a_i$ 从小到大排序之后，计算出值域，检查它们能不能生成 $[1, n]$ 中剩余的值即可。

// Date: Wed Dec 13 22:08:34 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 100010;
int t, n, a[N], b[N], c[N];
bool vis[N];

int main(void) {
#ifdef _DEBUG
  freopen("1617c.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n;

    For1(i, 1, n) { cin >> a[i]; }

    int len = 0;
    memset(vis, false, sizeof vis);

    For1(i, 1, n) {
      if (a[i] >= 1 && a[i] <= n && !vis[a[i]]) {
        vis[a[i]] = true;
      } else {
        b[++len] = a[i];
      }
    }

    int len1 = 0;
    For1(i, 1, n) {
      if (!vis[i]) {
        c[++len1] = i;
      }
    }

    sort(b + 1, b + 1 + len);
    bool flag = true;

    For1(i, 1, len) {
      int top = (b[i] - 1) / 2;
      if (top < c[i]) {
        flag = false;
        break;
      }
    }

    if (flag)
      cout << len << '\n';
    else
      cout << "-1\n";
  }

  return 0;
}