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
import collections
import heapq
import queue
import itertools
import functools
import operator
import time

readline = sys.stdin.readline
 
IPS = lambda: readline().rstrip()
IP = lambda: int(readline().rstrip())
MP = lambda: map(int, readline().split())
LS = lambda: list(map(int, readline().split()))

def solve():
    n = IP()
    arr = LS()
    ans,st = [],set()
    
    if arr[0]!=0: ans.append(0)
    else: ans.append(1)
    
    st.add(arr[0])
    
    for i in range(1,n):
        st.add(arr[i])
        if ans[-1] not in st: ans.append(ans[-1])
        else:
            j = 0
            while j<n+1:
                if j not in st:
                    ans.append(j)
                    break
                j+=1
            
    for elem in ans: print(elem, end=" ")
    print()


if __name__ == "__main__":
    for _ in range(1):
        solve()