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
import math as mt
input=sys.stdin.buffer.readline  
#t=int(input())
t=1
for __ in range(t):
    #n,x=map(int,input().split())
    n=int(input())
    l=list(map(int,input().split()))
    b=[-1]*n
    l1=[]
    for i in range(1,n):
        if l[i-1]!=l[i]:
            l1.append([l[i-1],l[i]])
    ch=0
    l2=[]
    for i in range(len(l1)):
        for j in range(l1[i][0]+1,l1[i][1]):
            l2.append(j)
    b=[]
    j=0
    if l[0]==0:
        if j<len(l2):
            b.append(l2[j])
            j+=1
        else:
            b.append(n+1)
    else:
        b.append(0)

    for i in range(1,n):
        if l[i]!=l[i-1]:
            b.append(l[i-1])
        else:
            if j<len(l2):
                b.append(l2[j])
            else:
                b.append(n+1)
            j+=1
    print(*b)        