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


// Problem : C. Paprika and Permutation
// Contest : Codeforces - Codeforces Round #761 (Div. 2)
// URL : https://codeforces.com/contest/1617/problem/C
// Memory Limit : 256 MB
// Time Limit : 1000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include<bits/stdc++.h>
#include<set>
#define lli long long int
#define pb push_back
#define rep(i,a,n) for(lli i=a;i<n;i++)
#define rrep(i,n,a) for(lli i=n-1;i>=a;i--) 
#define all(v) (v.begin(),v.end())
#define ppb pop_back
#define sz(v) v.size()
#define pi 3.141592653589793238
#define fr first
#define sc second
#define ins insert
const lli mod=1e9+7;
const int N = 1e5 + 5;
using namespace std;
lli lcm(lli a,lli b) { return (a/__gcd(a,b))*b;}
lli min(lli a,lli b,lli c) {return min(a,min(b,c));}
lli max(lli a,lli b,lli c) {return max(a,max(b,c));}
void solve(){
	lli n,a,b,c,k;
	cin>>n;
	lli ar[n];
	rep(i,0,n) cin>>ar[i];
	sort(ar,ar+n);
	bool ok=true;
	set<lli>st;
	rep(i,0,n) st.insert(i+1);
	
	
	
	lli ans=0;
	
	rep(i,0,n){
		if(ar[i]<=n) st.erase(ar[i]);
	}
	rep(i,0,n){

			k=*st.begin();
			if(k==ar[i]) st.erase(ar[i]);
			else if(k<(ar[i]+1)/2){
				st.erase(k);
				ar[i]=k;
				ans++;
			}
		
	}
	
	if(st.size()==0 and ok) cout<<ans<<endl;
	else cout<<"-1\n";
	
	
}
int main()
{

	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);	
	lli t=1;
	cin>>t;
	while(t--){
		solve();
	}

	
	
	return 0;
}