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.
#include<bits\stdc++.h>
#include<cstdio>
using namespace std;

int t,n,x,p,sum;
int num[1000000];
bool vis[1000000],is[1000005],a[1000000];
priority_queue<int> q,k;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;++i){
        scanf("%d",&a[i]);
        vis[a[i]]=true;
    }
    for(int i=0;i<=n;++i)
        if(!vis[i])     k.push(-i);
    for(int i=0;i<=n;++i)
    	q.push(-i);
    for(int i=1;i<=n;++i){
        while(q.size()&&is[-q.top()])
        	q.pop();
		if(-q.top()==a[i]){
            if(!k.empty()){
                num[i]=-k.top();
                vis[-k.top()]=is[-k.top()]=true;
                k.pop();
            }
            else
                num[i]=n+1,is[n+1]=true;;
            continue;
        }
        int z=-q.top();
        q.pop();
        while(q.size()&&is[-q.top()])
			q.pop();
        if(-q.top()!=a[i]){
            puts("-1");
			return 0;
        }
        num[i]=z, is[z]=true;
        vis[z]=true;
        if(-k.top()==z)
			k.pop();
    }
    for(int i=1;i<=n;++i)
    	printf("%d ",num[i]);
    return 0;
}