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())
l=list(map(int,input().split()))
from collections import Counter,defaultdict
if l[0]>1:
    print(-1)
else:
    ans=True
    l2=[]
    i=0
    d=defaultdict(int)
    l1=[0]*n
    l3=[]
    l4=[]
    x=0
    y=0
    for i in range(n):
        if l[i]>i+1:
            ans=False
        if d[l[i]]==0:
            if i==0 and l[i]==0:
                l3.append(0)
            elif i==0 and l[i]==1:
                l1[i]=0
            else:
                l1[i]=l[i-1]
            l2.append(l[i])
            d[l[i]]+=1
        else:
            l3.append(i)
    for i in range(1,n+2):
        if d[i]==0:
            l4.append(i)
    j=0
    for i in l3:
        if j!=len(l4):
            l1[i]=l4[j]
            j+=1
        else:
            l1[i]=l4[-1]
        
    if ans==True:
        print(*l1)
    else:
        print(-1)
        
                
     
