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=input().split()
li=[int(i) for i in l]
lfi=[0]
for i in li:
    lfi.append(i)
lx=[]
req=["N","N"]
done=1
for i in range(n,0,-1):
    
    if(lfi[i]>lfi[i-1]):
        if(req==["N","N"]):
            lx.append(lfi[i-1])
            req=[lfi[i-1]+1,lfi[i]-1]
            continue
        if(req[1]>=req[0]):
            done=0
            break
        lx.append(lfi[i-1])
        req=[lfi[i-1]+1,lfi[i]-1]
    else:
        if(req==["N","N"]):
            lx.append(10**6)
            continue
        if(req[1]<req[0]):
            lx.append(10**6)
            continue
        lx.append(req[1])
        if(req[1]>=req[0]):
            req[1]-=1

if(done==0):
    print(-1)
else:
    lx.reverse()
    for i in lx:
        print(i,end=" ")
    print()