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 math
import sys
from collections import deque
input = sys.stdin.readline


n = int(input())
a = list(map(int, input().split()))
if a[0] > 1:
    print(-1)
else:
    q = deque()
    for i in range(1, n):
        z = a[i] - a[i-1]
        if z > 1:
            for j in range(1, z):
                q.append(a[i-1] + j)
    ans = []
    k = 1
    if a[0] == 1:
        ans.append(0)
    else:
        if q:
            ans.append(q.popleft())
        else:
            ans.append(100001)
    for i in range(1, n):
        if a[i] != a[i-1]:
            ans.append(a[i-1])
        else:
            if q:
                ans.append(q.popleft())
            else:
                ans.append(100001)
    print(*ans)
