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.
#!/usr/bin/env python3
import sys
input = sys.stdin.readline

n = int(input())
a = [0] + [int(item) for item in input().split()]
ans = [-1] * n
used = set()
for i, (a1, a2) in enumerate(zip(a, a[1:])):
    if a1 > a2:
        print(-1)
        exit()
    if a1 < a2:
        ans[i] = a1
        used.add(a1)
val = a[-1] - 1
max_val = a[-1] + 1
for i in range(n-1, -1, -1):
    if ans[i] != -1:
        continue
    while val in used:
        val -= 1
    if val >= 0:
        ans[i] = val
    else:
        ans[i] = max_val
    used.add(val)
while val in used:
    val -= 1
if val >= 0:
    print(-1)
else:
    print(" ".join([str(item) for item in ans]))