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.
# cook your dish here
#code
import math
import collections
from sys import stdin,stdout,setrecursionlimit
from bisect import bisect_left as bsl
from bisect import bisect_right as bsr
import heapq as hq
setrecursionlimit(2**20)

t = 1
#t = int(stdin.readline())

for _ in range(t):
    n = int(stdin.readline())
    a = list(map(int, stdin.readline().rstrip().split()))
    
    n = len(a)
    b = []
    chk = False
    
    if(a[0]>1):
        chk = True
    
    if(a[0]==0):
        b.append(1)
    else:
        b.append(0)
    
    d = {b[0]:True}
    
    arr = [-1]
    
    for i in range(1,n):
        if(a[i]-a[i-1]==2 and (arr[-1]==2 or arr[-1]==1)):
            chk = True
            break
        if(a[i]>i+1 or a[i] in d):
            chk = True
            break
        else:
            if(a[i]==a[i-1]+2):
                if(a[i]-2 not in d):
                    b[-1] = a[i]-1
                    b.append(a[i]-2)
                    d[a[i]-2] = True
                else:
                    b.append(a[i]-1)
                    d[a[i]-1] = True
            elif(a[i]==a[i-1]):
                tmp = b[-1]
                b.append(tmp)
                d[tmp] = True
            elif(a[i]==a[i-1]+1):
                b.append(a[i]-1)
                d[a[i]-1] = True
            else:
                chk = True
                break
        arr.append(a[i]-a[i-1])
            
            
    if(chk):
        print(-1)
        continue
    else:
        for i in b:
            print(i,end=' ')
        print()