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 io
import os

from collections import Counter, defaultdict, deque
import heapq

BIG = 10**6
def solve(N, A):
    B = [
        None for i in range(N)
    ]
    need = []
    for i in range(N - 2, -1, -1):
        x = A[i]
        y = A[i + 1]
        if x < y:
            B[i + 1] = x
            for z in range(x + 1, y):
                heapq.heappush(need, z)
        else:
            if need:
                B[i + 1] = heapq.heappop(need)
            else:
                B[i + 1] = BIG
    if 0 < A[0]:
        if A[0] != 1:
            return -1
        B[0] = 0
    else:
        if len(need) > 1:
            return -1
        if len(need) == 1:
            B[0] = need[0]
        else:
            B[0] = BIG
    
    return ' '.join(map(str, B))


if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    N, = [int(x) for x in input().split()]
    A = [int(x) for x in input().split()]
    ans = solve(N, A)
    print(ans)
