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())
nums = [int(x) for x in input().split()]
stack = []
ans = [-1]*n
for i in range(n-1,0,-1):
    if nums[i]>i+1:
        print(-1)
        break
    if nums[i]==nums[i-1]:
        if len(stack)>0:
            ans[i]=stack.pop()
        else:
            ans[i]=1000000
    else:
        ans[i]=nums[i-1]
        delta = nums[i]-nums[i-1]
        start = nums[i-1]+1
        for j in range(delta-1):
            stack.append(start)
            start+=1
        

if nums[0]==0:
    if len(stack)>0:
        ans[0]=stack.pop()
    else:
        ans[0]=1000000
elif nums[0]==1:
    ans[0]=0

for i in range(n):
    print(ans[i],end=" ")
