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.
import sys
input=sys.stdin.buffer.readline

#strategy: if i is a changing point i.e. a[i]!=a[i-1], then set b[i]=a[i-1]
#if i is not a changing point, store the next largest value into b[i] that isn't in a

n=int(input())
A=[int(x) for x in input().split()]


B=[0 for _ in range(n)]

aValues=set(A) #store unique a values
prevVal=0
#currI=-1
possible=True

nextValue=1
while nextValue in aValues:
    nextValue+=1
    
for i in range(n):
    if A[i]!=prevVal: #changing point
        if A[i]-1>i: #not enough indexes to fill the gaps
            possible=False
            break
        B[i]=prevVal
        prevVal=A[i]
    else: #non-changing point
        B[i]=nextValue
        nextValue+=1
        while nextValue in aValues:
            nextValue+=1
        
if possible==True:
    print(' '.join([str(x) for x in B]))
else:
    print('-1')