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()))
x=[]
m=[]
p=[]
ans=[]
flag=0
if a[0]>1:
    flag=1
else:
    for i in range(1,n):
        if a[i]>i+1:
            flag=1
            break
        else:
            d=a[i]-a[i-1]
            x.append(d)
            if d>1:
                m.append(d-1)
                p.append(a[i])
    if a[0]==1:
        ans.append(0)
    else:
        if len(p)!=0:
            ans.append(p[0]-m[0])
            m[0]-=1
            if m[0]==0:
                m.pop(0)
                p.pop(0)
        else:
            ans.append(a[n-1]+1)
    for i in range(1,n):
        if x[i-1]>0:
            ans.append(a[i-1])
        else:
            if len(p) != 0:
                ans.append(p[0] - m[0])
                m[0] -= 1
                if m[0] == 0:
                    m.pop(0)
                    p.pop(0)
            else:
                ans.append(a[n-1]+1)
if flag==1:
    print(-1)
else:
    for i in range(len(ans)):
        print(ans[i],end=" ")
