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.readline

n = int(input())
arr = list(map(int ,input().split()))

vis =[0]*(n+1)    # 0<= a[i] <=i
b = [-1]*n
for i in range(1, n) :
    if arr[i] != arr[i-1] :
        vis[arr[i-1]] = 1
        b[i] = arr[i-1]

k = 0
vis[arr[-1]] = 1 #max of all number which appeared

for i in range(n) :
    if b[i] == -1 :
        # to check 0 or 1 and first minimum which is not appeared
        while vis[k] == 1 :
            k +=1
        b[i] = k
        k +=1

print(*b)       




        
                
                  
                
            
            
    
    
        
