human
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.
Solve the task in PYTHON3.
 ... Can you provide a Python code to determine the minimum number of operations needed to make the array a permutation of integers 1 to n?
assistant
#Here is Python code to determine the minimum number of operations needed to make the array a permutation of integers 1 to n.

import array
import bisect
import heapq
import math
import collections
import sys
import copy
from functools import reduce
import decimal
from io import BytesIO, IOBase
import os
import itertools
import functools
from types import GeneratorType

#
# sys.setrecursionlimit(10 ** 9)
decimal.getcontext().rounding = decimal.ROUND_HALF_UP

graphDict = collections.defaultdict

queue = collections.deque

################## pypy deep recursion handling ##############
# Author = @pajenegod
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        to = f(*args, **kwargs)
        if stack:
            return to
        else:
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        return to
                    to = stack[-1].send(to)

    return wrappedfunc

################## Graphs ###################
class Graphs:
    def __init__(self):
        self.graph = graphDict(set)

    def add_edge(self, u, v):
        self.graph[u].add(v)
        self.graph[v].add(u)

    def dfs_utility(self, nodes, visited_nodes, colors, parity, level):
        global count
        if nodes == 1:
            colors[nodes] = -1
        else:
            if len(self.graph[nodes]) == 1 and parity % 2 == 0:
                if q == 1:
                    colors[nodes] = 1
                else:
                    colors[nodes] = -1
                    count += 1
            else:
                if parity % 2 == 0:
                    colors[nodes] = -1
                else:
                    colors[nodes] = 1
        visited_nodes.add(nodes)
        for neighbour in self.graph[nodes]:
            new_level = level + 1
            if neighbour not in visited_nodes:
                self.dfs_utility(neighbour, visited_nodes, colors, level - 1, new_level)

    def dfs(self, node):
        Visited = set()
        color =...