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.
from collections import defaultdict
n=int(input())
l=list(map(int,input().split()))
l1=[]
d=defaultdict(int)
for x in l:
    d[x]=1
ma=1
if l[0]==0:
    while(d[ma]!=0):
            ma+=1
    d[ma]=1
    l1.append(ma)
    m=0
elif l[0]==1:
    l1.append(0)
    m=1
    ma=2
else:
    f=1
    sys.exit(0)
f=0
for x in range(1,n):
    if l[x]==m:
        while(d[ma]!=0):
            ma+=1
        d[ma]=1
        l1.append(ma)
    else:
        if l[x]>ma+1:
            f=1
            break
        l1.append(m)
        m=l[x]
        ma=max(ma,m)
if f==1:
    print(-1)
else:
    for x in l1:
        print(x,end=" ")
    
