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.
m=int(input())

a=[]
b={-1}
c={-1}
c.pop()
a=input().split()
def process(n):
    rem=-1
    b.pop()
    for j in range(0,n-1):
        b.add(j)
        if int(a[j])>j+2:
            print("-1")
            return
    if int(a[n-1])>=n+1:
        print("-1")
        return

    b.add(n-1)
    i=0
    while i!=n:
        if rem!=-1:
            c.add(rem)
            rem=-1
        if int(a[i]) in b:
            b.remove(int(a[i]))
            rem=int(a[i])
        i=i+1
    return rem



rem=process(m)


if rem!=-1:
    for x in b:
        print(x,end=' ')
        break
for x in b:
    print(x,end=' ')
for x in c:
    print(x,end=' ')
