Given an array a of length n, find another array, b, of length n such that:

  * for each i (1 ≤ i ≤ n) MEX(\\{b_1, b_2, …, b_i\})=a_i. 



The MEX of a set of integers is the smallest non-negative integer that doesn't belong to this set.

If such array doesn't exist, determine this.

Input

The first line contains an integer n (1 ≤ n ≤ 10^5) — the length of the array a.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ i) — the elements of the array a. It's guaranteed that a_i ≤ a_{i+1} for 1≤ i < n.

Output

If there's no such array, print a single line containing -1.

Otherwise, print a single line containing n integers b_1, b_2, …, b_n (0 ≤ b_i ≤ 10^6)

If there are multiple answers, print any.

Examples

Input


3
1 2 3


Output


0 1 2 

Input


4
0 0 0 2


Output


1 3 4 0 

Input


3
1 1 3


Output


0 2 1 

Note

In the second test case, other answers like [1,1,1,0], for example, are valid.
n=int(input())
a=list(map(int,input().split()))
b=[-1]*n
if a[0]==0 or a[0]==1:
    maxi=-1
    if a[0]==1:
        b[0]=0
        maxi=0
    for i in range(1,n):
        if a[i]!=a[i-1]:
            b[i]=a[i-1]
    tofill=[]
    i=0
    while i<n:
        if b[i]==-1:
            tofill.append(i)
        elif b[i]!=-1 and len(tofill)!=0:
            maxi=max(maxi,a[i-1])
            for j in range(len(tofill)):
                if maxi!=a[i]-1:
                    b[tofill[0]]=maxi+1
                    tofill.pop(0)
                    maxi+=1
                else:
                    break
            if maxi!=a[i]-1:
                print(-1)
                break
        i+=1
    if len(tofill)==0 and b.count(-1)!=0:
        pass
    else:
        maxi=a[-1]
        for i in range(len(tofill)):
            b[tofill[i]]=maxi+1
            maxi+=1
        print(*b)
else:
    print(-1)
