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 findmiss(arr, n):
    for i in range(n):
        if (arr[i] <= 0 or arr[i] > n):
            continue

        val = arr[i]

        while (arr[val - 1] != val):
            nextval = arr[val - 1]
            arr[val - 1] = val
            val = nextval
            if (val <= 0 or val > n):
                break

    for i in range(n):
        if (arr[i] != i + 1):
            return i + 1

    return n + 1


n=int(input())
mt = list(map(int, input().split()))
ans=[]
for i in range (n):
    if(mt[i]>(i+1)):
        print("-1")
        break
else:
    for i in range(n):
        ans.append(i)
    if (mt[n - 1] in ans):
        id = ans.index(mt[n - 1])
        ans[id] = n
    for i in range(n-1):
        tp = n-i-2
        if(mt[tp] in ans[0:tp+1]):
            ans[tp+1]=mt[tp]
            id= ans.index(mt[tp])
            form = ans[0:tp+1] + mt[id:]
            m= findmiss(form,len(form))
            ans[id]=m
    print(ans)