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())
czy=0
T,A,N,P=[0],[0],[0],[0]
T+=map(int,input().split())
for i in range (1,n+1):
    A.append(-1)
    if T[i]>i:
        czy=1;
    N.append(i+1)
    P.append(i-1)
N.append(0)
P.append(0)
if czy:
    print(-1)
    exit()
for i in range(1,n+1):
    x=T[i]-T[i-1]
    ile=T[i-1]
    s=i
    for j in range(1,x+1):
        A[s]=ile
        ile+=1
        P[N[s]]=P[s]
        N[P[s]]=N[s]
        s=P[s]    
for i in range(1,n+1):
    if A[i] == -1:
        A[i]=1000000
s=""
for i in range(1,n+1):
    s+=str(A[i])+" "
print(s)