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()))
l=[0]*(1000001)
for i in a:
    l[i]=1
s,np,f,j,k=[],[],[],0,0
for i in  range(0,1000001):
    if l[i]==0:
        np.append(i)
w=len(np)        
fg=0    
for i in range(0,n):
    if a[i]>(i+1):
        fg=1
        break
if fg==1:
    print(-1)
else:
    for i in range(0,n):
        b=len(s)
        if j<b and k<w:
            if s[j]<np[k] and s[j]!=a[i]:
                f.append(s[j])
                j+=1
            else:
                f.append(np[k])
                k+=1
        elif k>=w and j<b:
            if s[j]==a[i]:
                fg=1
                break
            else:
                f.append(s[j])
                j+=1
        elif k<w and j>=b:
            f.append(np[k])
            k+=1
        else:
            fg=1
            break
        if b!=0:
            if s[b-1]!=a[i]:
                s.append(a[i])
        else:
            s.append(a[i])
if fg==1:
    print(-1)
else:
    print(" ".join(str(a) for a in f))