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.
def main():
    n = int(input())
    a = list(map(int, input().split()))
    b = []
    need = a[-1] - 1; done = set()
    million = 10 ** 6
    for i in range(n - 1, 0, -1):
        if a[i] == a[i - 1]:
            while True:
                if need in done:
                    need -= 1
                elif need < a[i]:
                    b.append(million)
                    break
                else:
                    b.append(need)
                    done.add(need)
                    break
        else:
            b.append(a[i - 1])
            done.add(a[i - 1])
        #print(b)
    if a[0] == 0:
        while True:
            if need in done:
                need -= 1
            elif need < a[0]:
                b.append(million)
                break
            else:
                b.append(need)
                done.add(need)
                break
    else:
        b.append(0)
    b.reverse()
    print(*b)
main()