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>
#define ll long long
using namespace std;

void solve(){
	int n;
	cin >> n;
	int a[n], b[n] = {};
	set<int> s;
	for(int i = 0; i < n; i++){
		cin >> a[i];
		s.insert(a[i]);
	}
	int tmp = 0;
	if(a[0] == 0){
		tmp = 1;
		b[0] = 1;
	}
	else{
		b[0] = tmp;
	}
	for(int i = 1; i < n; i++){
		if(a[i-1] != a[i]){
			b[i] = *s.begin();
			s.erase(b[i]);
		}
		else {
			tmp++;
			while(s.count(tmp)){
				tmp++;
			}
			b[i] = tmp;
		}
	}
	for(int i = 0; i < n; i++){
		cout << b[i] << ' ';
	}
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int t = 1;
	//cin >> t;
	while(t--) solve(); 
}