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>
using namespace std;

#define int long long
#define endl "\n"
#define mod 1000000007
#define pb push_back
#define ff first
#define ss second
#define FIO ios_base::sync_with_stdio(false); cin.tie(NULL);


main(){
	FIO
	int n;
	cin>>n;
	int arr[n],i,j,k;
	set<int>s;
	for(i=0;i<n;i++){
	    cin>>arr[i];
	    s.insert(arr[i]);
	}
	bool f=0;
	if(arr[0]>1){
	    cout<<-1;
	    return 0;
	}
	int brr[n];
	set<int>ss;
	for(i=0;i<n;i++){
	    if(arr[i]>i+1||arr[i]<arr[i-1]||ss.find(arr[i])!=ss.end()){
	        cout<<-1;
	        return 0;
	    }
	    if(arr[i]==arr[i-1]){
	        for(k=j;;k++){
	            if(s.find(k)==s.end()){
	                j = k+1;
	                brr[i] = k;
	                break;
	            }
	        }
	    }
	    else{
	        brr[i] = arr[i-1];
	        ss.insert(brr[i]);
	    }
	}
	for(i=0;i<n;i++){
	    cout<<brr[i]<<" ";
	}
}