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

n = int(sys.stdin.readline().strip())
a = list(map(int, sys.stdin.readline().strip().split()))
v = True
for i in range (0, n):
    if a[i] > i + 1:
        v = False
    if i > 0:
        if a[i] < a[i-1]:
            v = False
if v == False:
    print(-1)
else:
    c = []
    i = 0
    j = 0
    while i < n:
        if a[i] == j:
            i = i + 1
            j = j + 1
        elif j < a[i]:
            c.append(j)
            j = j + 1
        else:
            i = i + 1
    c = c + [(a[n-1]+1) for i in range (0, n)]
    j = 0
    b = []
    for i in range (0, n):
        if i > 0:
            if a[i-1] < a[i]:
                b.append(a[i-1])
            else:
                b.append(c[j])
                j = j + 1
        else:
            b.append(c[j])
            j = j + 1
    print(" ".join(list(map(str, b))))

    
    
