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;
#define ll long long int
#define pii pair<ll,ll>
#define rep(i,st,en) for(ll i=st;i<en;i++)
#define rrep(i,st,en) for(ll i=en-1;i>=st;i--)
#define vi vector<ll>
#define vii vector<pii>
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define eb emplace_back
#define yes cout<<"YES"<<endl; return;
#define no cout<<"NO"<<endl; return;
#define flus fflush(stdout);
#define debug(x) cerr<<#x<<" : "<<x<<" ";
#define fin cerr<<endl;
#define ff first
#define ss second
#define sz size
#define pb push_back
#define np next_permutation
#define inf 1e18
#define Mod 1000000007
#define set_bits __builtin_popcountll
#define PI 3.14159265358979323846
#define minpq priority_queue <ll, vector<ll>, greater<ll>>
#define maxpq priority_queue<ll> 
#define nline '\n'
#define endl '\n'
vector<ll> sieve(int n) {int*arr = new int[n + 1](); vector<ll> vect; for (int i = 2; i <= n; i++)if (arr[i] == 0) {vect.push_back(i); for (int j = 2 * i; j <= n; j += i)arr[j] = 1;} return vect;}

void solve(){
    int n;
    cin>>n;

    int a[n];
        
    vector<int> temp;
    int b[n]{0};
    rep(i,0,n){
        cin>>a[i];

        if(b[i]==0 && a[i]<=n){
            b[a[i]-1]=a[i];
        }else{
            temp.pb(a[i]);
        }
    }

    sort(all(temp));

    int p1=0;
    rep(i,0,n){
        if(b[i]==0){
            b[i]=temp[p1];
            p1++;
        }
    }

    rep(i,0,n){
        a[i]=b[i];
        // cout<<a[i]<<" ";
    }
    // cout<<endl;

    int ans =0;

    rep(i,0,n){
        if(a[i]==i+1){
            continue;
        }

        if((a[i]%(a[i]-(i+1))==i+1)){
            ans++;
        }else{
            cout<<-1<<endl;
            return;
        }
    }
    cout<<ans<<endl;

}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    clock_t t1=clock();
    int t=1;
    cin>>t;  

    while(t--){ 
        solve();
    }
    cerr<<"Time elapsed: "<<(double)(clock()-t1)/1000<<" s"<<endl;
}
