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
#input = sys.stdin.readline
def main():
    n = int( input())
    A = list( map( int, input().split()))
    B = [-1]*n
    N = 10**6
    now = N
    TB = [True]*N
    if A[0] > 0:
        B[0] = 0
        TB[0] = False
    for i in range(1,n):
        a = A[i]
        ab = A[i-1]
        if ab < a:
            if TB[ab]:
                B[i] = ab
                TB[ab] = False
    TB[A[-1]] = False
    indexb = 0
    for i in range(n):
        if B[i] != -1:
            continue
        for j in range(indexb, N):
            if TB[j]:
                TB[j] = False
                indexb = j+1
                B[i] = j
                break
    indexa = 0
    TA = [True]*N
    for i in range(n):
        TA[B[i]] = False
        for j in range(indexa, N):
            if TA[j]:
                if A[i] != j:
                    print(-1)
                    return
                indexa = j
                break
        
    print(" ".join( map( str, B)))
        
    
if __name__ == '__main__':
    main()
