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 gc getchar_unlocked
#define fo(i, n) for (i = 0; i < n; i++)
#define Fo(i, k, n) for (i = k; k < n ? i < n : i > n; k < n ? i += 1 : i -= 1)
#define ll long long
#define si(x) scanf("%d", &x)
#define sl(x) scanf("%lld", &x)
#define ss(s) scanf("%s", s)
#define pi(x) printf("%d\n", x)
#define pl(x) printf("%lld\n", x)
#define ps(s) printf("%s\n", s)
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for (auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
typedef pair<int, int> pii;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpii;
typedef vector<pl> vpl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim)
{
    uniform_int_distribution<int> uid(0, lim - 1);
    return uid(rang);
}
const int mod = 1'000'000'007;
ll inv(ll i) {if (i == 1) return 1; return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;}
 
ll mod_mul(ll a, ll b) {a = a % mod; b = b % mod; return (((a * b) % mod) + mod) % mod;}
 
ll mod_add(ll a, ll b) {a = a % mod; b = b % mod; return (((a + b) % mod) + mod) % mod;}
 
ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b);}
 
ll ceil_div(ll a, ll b) {return a % b == 0 ? a / b : a / b + 1;}
 
int mpow(int base, int exp)
{
    base %= mod;
    int result = 1;
    while (exp > 0)
    {
        if (exp & 1)
            result = ((ll)result * base) % mod;
        base = ((ll)base * base) % mod;
        exp >>= 1;
    }
    return result;
}
/*
const int N=1e5+1;
int a[N],tree[4*N];
void construct_maxSegTree(int node,int st,int end){
    if(st==end) {
        tree[node]=a[st];
        return;
    }
    int mid=(st+end)/2;
    construct_maxSegTree(2*node,st,mid);
    construct_maxSegTree(2*node+1,mid+1,end);
    tree[node]=max(tree[2*node],tree[2*node+1]);
}
int queryMax(int node,int st,int end,int l,int r){
  if(end<l || st>r) return INT_MIN;
   if(st>=l && end<=r) return tree[node];
   int mid=(st+end)/2;
   int q1=queryMax(2*node, st, mid,l,r);
   int q2=queryMax(2*node+1,mid+1,end,l,r);
   return max(q1,q2);
}
void updateMax(int node,int st,int end,int ind,int val){
   if(st==end){
       a[st]=val;
       tree[node]=val;
       return;
   }
   int mid=(st+end)/2;
   if(ind<=mid){
       updateMax(2*node,st,mid,ind,val);
   }
   else if(ind>mid){
       updateMax(2*node+1,mid+1,end,ind,val);
   }
   tree[node]=max(tree[2*node],tree[2*node+1]);
}*/
string DecimalToBinaryString(int a)
{
    string binary = "";
    int mask = 1;
    for(int i = 0; i < 31; i++)
    {
        if((mask&a) >= 1)
            binary = "1"+binary;
        else
            binary = "0"+binary;
        mask<<=1;
    }
    cout<<binary<<endl;
    return binary;
}
//#@copyrightMUKESH
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////*
void solve()
{
 int n;
 cin>>n;
 ll a[n];
 for(int i=1;i<=n;i++){
     cin>>a[i];
 }
 sort(a+1,a+n+1);
 //deb(a[9]);
 int cnt=0;
 for(int i=1;i<=n;i++){
     if(a[i]!=i){
    if(a[i]%(a[i]-i)!=0 ){
        cnt++;
    }
    else if(a[i]%(a[i]-i)==0 ){
        cout<<-1<<"\n";
        //deb(i);
        return;
    }
     }
 }
cout<<cnt<<"\n";
}

int main()
{
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    srand(chrono::high_resolution_clock::now().time_since_epoch().count());

    int t = 1;
     cin >> t;
    
    while (t--)
    {
       solve();
    }

    return 0;
}
