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.
#BECAUSE ONE TRUTH PREVAILS
import sys
input=sys.stdin.readline
from collections import defaultdict
N=int(input())
L=list(map(int,input().split()))
ok=True
for i in range(N):
    if L[i]>i+1:
        ok=False
        break
if (ok==False):
    print(-1)
else:
    Stack=[]
    ans=[0]*N
    for i in range(N):
        if(len(Stack)==0):
            Stack.append(i)
        else:
            if L[i]==L[Stack[-1]]:
                continue
            else:
                index2=i
                index1=Stack[-1]
                Stack.pop()
                t=L[index1]-1
                for k in range(index1,index2+1):
                    ans[k]=t
                    t+=1
                    if t>=L[index2]:
                        t=L[index1]
                Stack.append(i)
    print(*ans)

