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.
"""T=int(input())
for _ in range(0,T):
    n=int(input())
    a,b=map(int,input().split())
    s=input()
    s=[int(x) for x in input().split()]
    for i in range(0,len(s)):
        a,b=map(int,input().split())"""


from collections import deque

n=int(input())
s=[int(x) for x in input().split()]
temp=0
for i in range(0,len(s)):
    if(s[i]>i+1):
        temp=1
        break

if(temp==1):
    print(-1)
else:
    pos=[0]*((10*n)+5)
    for i in range(0,len(s)):
        pos[s[i]]=1


    b=[]
    ptr=0
    kk=0
    dd=dict()
    for i in range(0,len(s)):
        if(ptr==s[i]):
            for j in range(kk,len(pos)):
                if(pos[j]==0):
                    b.append(j)
                    pos[j]=1
                    dd[j]=1
                    kk=j+1
                    break

        else:
            b.append(ptr)
            dd[ptr]=1
            pos[ptr]=1
            for j in range(ptr+1,len(pos)):
                if(dd.get(j)==None):
                    ptr=j
                    break
            



    
    print(*b)
            
    

