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.
from sys import stdin, gettrace

if gettrace():
    inputi = input
else:
    def input():
        return next(stdin)[:-1]


    def inputi():
        return stdin.buffer.readline()


def main():
    n = int(inputi())
    aa = [int(a) for a in inputi().split()]
    bb = [0] * n
    ap = 0
    bv = 0
    mex = 0
    ap = 0
    bv = 0
    for i in range(n):
        if i > 0 and aa[i] != aa[i-1]:
            bb[i] = aa[i-1]
            if aa[i] > max(bv, bb[i] + 1):
                print(-1)
                return
            bv = max(bb[i]+1, bv)
        else:
            while ap < n and bv >= aa[ap]:
                if bv == aa[ap]:
                    bv += 1
                ap += 1
            bb[i] = bv
            bv += 1
            if aa[i] > bv + 1:
                print(-1)
                return
    print(' '.join(map(str, bb)))




if __name__ == "__main__":
    main()