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
from fractions import gcd
from math import ceil,sqrt,sin,pi,tan
from collections import defaultdict
read=sys.stdin.buffer.readline
import heapq
mi=lambda:map(int,read().split())
li=lambda:list(mi())
cin=lambda:int(read())

for _t in range(1):
    n=cin()
    a=li()
    mex=[-1]*100005
    li=a[0]
    for i in range(1,n):
        if li!=a[i]:
            mex[li]=a[i]
            li=a[i]
    mex[li]=-2
    li=0
    for i in range(100005):
        if mex[i]==-1:
            li=i
            break
    x=li
    for i in range(li+1,100005):
        if mex[i]==-1:
            mex[li]=i
            li=i
    mex[li]=-2
    ans=[x]
    le=a[0]
    y=a[0]
    x=mex[x]
    
    if a[0]>1:
        print(-1)
        exit(0)
    for i in range(1,n):
        if a[i]!=le:
            le=a[i]
            ans.append(y)
            y=mex[y]
        else:
            ans.append(x)
            x=mex[x]
        if min(y,x)!=a[i]:
            print(-1)
            exit(0)
    print(*ans)