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

import sys
import os.path
 
if(os.path.exists('input.txt')) :
    sys.stdin = open("input.txt", "r")
    sys.stdout = open("output.txt", "w")
    sys.stderr = open("error.txt", "w")
    
depth = 200001
mod = 1000000007 
lim = mod * mod
sys.setrecursionlimit(depth) 
 
linp = lambda: list(minp())
minp = lambda: map(int, input().split())
 
from math import inf, ceil, sqrt, log2, gcd
from collections import defaultdict, deque

dxy = [(1, 0),(-1, 0),(0, 1),(0, -1)]

def expo(a, b) :
    if not b : return 1
    v = expo(a, b//2)
    v = (v * v) % mod
    if b%2 : v = (v * a) % mod
    return v

def modinv(a, b) :
    d = expo(b, mod-2)
    v = (a * d) % mod
    return v

def checkPrime(n) :
    if n < 4 : return True
    if n%2 == 0 : return False
    for i in range(3, int(sqrt(n))+1, 2) :
        if n%i == 0 :
            return False
    return True    

def highestPowerof2(x):
    x |= x >> 1
    x |= x >> 2
    x |= x >> 4
    x |= x >> 8
    x |= x >> 16

    return x ^ (x >> 1)

for _ in range(int(input())) :
    n = int(input())
    a = sorted(linp())
    b = [False for _ in range(n)]
    c = [True for _ in range(n)]

    for i in range(n) :
        if a[i] <= n and not b[a[i]-1] :
            c[i] = False 
            b[a[i]-1] = True 

    d = list()
    for i in range(n) :
        if c[i] :
            d.append(a[i])

    e = list()
    for i in range(n) :
        if not b[i] :
            e.append(i+1)

    a, b = d, e
    a.sort(), b.sort()
    
    n, f = len(a), True
    for i in range(n) :
        if b[i] >= (a[i]-b[i]) :
            f = False

    if f : print(n)
    else : print(-1)