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.
n = int(input())
A = list(map(int, input().split()))
z = True
for i in range(len(A)):
    if A[i] - A[0] > i:
        z = False
        break
if not z:
    print(-1)
else:
    C = []
    B = [-10]
    for i in range(1, len(A)):
        if A[i] != A[i - 1]:
            B.append(A[i - 1])
            C.append(A[i - 1])
        else:
            B.append(-10)
    C.append(A[-1])
    C.append(10 ** 18)
    q = 0
    X = 0
    #print(C)
    #print(B)
    for i in range(len(B)):
        if B[i] == -10:
            while X == C[q]:
                q += 1
                X += 1

            B[i] = X
            X += 1


    print(*B)