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;

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin >> n;
	vector<int> ar(n),ans;
	int cnt[n] = {0};
	set<int> st,tans;
	for(auto &id : ar){
		cin >> id;
		st.insert(id);
		cnt[id]++;
	}
	int lsBl = 0, mxBl = 0;
	for(auto &id : ar){
		if(mxBl >= id-2){
			while(st.count(lsBl)){
				lsBl++;
			}
			st.insert(lsBl);
			ans.push_back(lsBl);
			tans.insert(lsBl);
			mxBl = lsBl;
//			cout << lsBl << '\n';
			lsBl = id;
			if(cnt[id] == 1){
				st.erase(id);
			}
			cnt[id]--;
		}else{
			cout << "-1\n";
			return 0;
		}
	}
	int mex = 0;
	for(auto id : tans) if(mex == id) mex = id+1;
	if(mex != ar[ar.size()-1]){
		cout << "-1\n";
		return 0;
	}
	for(auto it : ans) cout << it << ' ';
	cout << '\n';
}