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()))
cnt = [0 for i in range(max(A)+1)]
for i in A:
    cnt[i]+=1
D = set(A);D=list(D)
s=1
for i in D:
    if(cnt[i] in D):
        s=1
        break
f=0
for i in range(1,n):
    if(A[i]-A[i-1]>2):
        f=1
        break
    if(A[i]>i+1):
        f=1
        break
if(f==1 or len(D)==1 or s==1):
    print("-1")
else:
    B=[]
    if(A[0]==0):
        B.append(1)
    else:
        B.append(0)
    for i in range(1,n):
        if(A[i]==0):
            B.append(1)
            continue
        C=[max(A[i]-2,0),max(A[i]-1,0),A[i]+1,A[i]+2]
        c=0
        for j in C:
            if(j not in B):
                B.append(j)
                c=1
                break
        if(c==0):
            break
    print(*B)
    
            
            