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

# cook your dish here
import sys
import bisect
from bisect import bisect_left as lb
from bisect import bisect_right as rb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
sflo=lambda:list(map(float,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
sys.setrecursionlimit(3*10**5)
mod=10**9+7
mod1=998244353
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs2(root,par):
    if par!=-1:
        dp[root]=dp[par]+1
    for i in range(1,20):
        if kthpar[root][i-1]!=-1:
            kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
    for child in gp[root]:
        if child==par:continue
        kthpar[child][0]=root
        dfs(child,root)
        
ans=0
a=[]
n,k=0,0
b=[]
vis=[]
tot=0
time=[]
time1=[]
adj=[]
mx=-1
eps=0.0000001
x=0
pref=[]
lvl=[]
gp=[]
ans=[]
f=0
w=[]

def dfs(i,prod,rem,n):
    if rem==0:
        while(i<len(n)):
            prod*=int(n[i])
            i+=1
        return prod
    if i==len(n):
        return prod
    mx=0
    for j in range(9-int(n[i])):
        mx=max(dfs(i+1,(int(n[i])+min(j,rem))*prod,rem-min(j,rem),n),mx)
    return mx
def hnbhai(tc):
    n=sb()
    a=sd()
    are=[]
    last=1
    a.sort()
    hogaya=[0]*(n+2)
    ans=0
    for i in a:
        if i>=1 and i<=n and hogaya[i]==0:
            hogaya[i]=1
            ans-=1
    d={}
    mark={}
    #print(hogaya)
    for i in a:
        d[i]=d.get(i,0)+1
        if mark.get(i)is None:
            d[i]-=hogaya[min(n+1,i)]
            mark[i]=1
    are=list(d.keys())
    are.sort()
    #print(are)
    last=1
    tot=0
    rem=0
    mn=[0]*(n+1)
    for i in range(len(are)-1,-1,-1):
        kaha=are[i]//2
        if are[i]%2==1:kaha+=0
        else:kaha-=1
        mn[i]=max(kaha,mn[i+1])
    t=0
    #print(d)
    for i in are:
        t+=1
        while(last<=n and hogaya[last]):
            last+=1
        ok=i if i>=1 and i<=n else -1
        kaha=i//2
        #print(i,last)
        if i%2==1:kaha+=0
        else:kaha-=1
        #print(i,d[i])
        while(last<=n and last<=kaha) and d[i]>0:
            if hogaya[last]:
                last+=1
                continue
            rem-=1
            d[i]-=1
            #print(last)
            last+=1
            tot+=1
        if d[i]>0:
            print("-1")
            return
    #print(d)
    while(last<=n and hogaya[last]==1):last+=1
    if last<n+1:
        print(-1)
        return
    #print(last,d)
    print(tot)
    return
for _ in range(sb()):
    hnbhai(_+1)
##case

#.
