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


def getint():
    return [int(i) for i in input().split()]
def get():
    return int(input())
def getstr():
    return [i for i in input().split()]
def S():
    for test in range(int(input())):
        solve()
import math
import itertools as it
import bisect
import time
import collections as ct

def lower_bound(a,x):
    l=-1
    r=len(a)
    while l+1!=r:
        mid=l+r>>1
        if a[mid]<x:
            l=mid
        else:
            r=mid
    return r
def upper_bound(a,x):
    l=-1
    r=len(a)
    while l+1!=r:
        mid=l+r>>1
        if a[mid]<=x:
            l=mid
        else:
            r=mid
    return r
def solve():
    n=get()
    a=getint()
    a.sort()
    useful=[]
    cha=[]
    s={}
    for i in range(n):
        if 1<=a[i]<=n and s.get(a[i],0)!=1:
            useful.append(a[i])
            s[a[i]]=1
    # print(useful)
    check=[]
    for i in range(n):
        if s.get(a[i],0)==0:
            check.append(a[i])
    for i in range(1,n+1):
        if s.get(i,0)==0:
            cha.append(i)
    # print(check)
    for i in range(len(check)):
        if check[i]&1:
            check[i]//=2
        else:
            check[i]=check[i]//2-1
    check.sort()
    cha.sort()
    if len(check)!=len(cha):
        print(-1)
        return
    for i in range(len(check)):
        if check[i]<cha[i]:
            print(-1)
            return
    print(len(check))

S()
