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()))
ans = True
b = []
f=False
f1  =False
if a[0]==1:
    b.append(0)
    mex = 1
    j = 2
    f1 = True
elif a[0]==0:
    b.append(1)
    mex = 0
    j = 3
else:
    ans = False
if ans:
    if f1:
        m = 3
        for i in range(1,n):
            k = a[i]
            if k==mex:
                b.append(j)
                j+=1
            elif k==m:
                b.append(mex)
                mex = m
                m = j
                j = j+1
            else:
                ans=False
                break
    else:
        m=2
        for i in range(1,n):
            k = a[i]
            if k==mex:
                b.append(j)
                j+=1
            elif k==m:
                b.append(mex)
                m = j
                j+=1
            else:
                ans=False
                break

if ans:
    for i in b:
        print(i,end=" ")
    print()
else:
    print(-1)