題幹：

Companies always have a lot of equipment, furniture and other things. All of them should be tracked. To do this, there is an inventory number assigned with each item. It is much easier to create a database by using those numbers and keep the track of everything.

During an audit, you were surprised to find out that the items are not numbered sequentially, and some items even share the same inventory number! There is an urgent need to fix it. You have chosen to make the numbers of the items sequential, starting with 1. Changing a number is quite a time-consuming process, and you would like to make maximum use of the current numbering.

You have been given information on current inventory numbers for n items in the company. Renumber items so that their inventory numbers form a permutation of numbers from 1 to n by changing the number of as few items as possible. Let us remind you that a set of n numbers forms a permutation if all the numbers are in the range from 1 to n

Input

The first line contains a single integer n — the number of items (1 ≤ n ≤ 105).

The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — the initial inventory numbers of the items.

Output

Print n numbers — the final inventory numbers of the items in the order they occur in the input. If there are multiple possible answers, you may print any of them.

Examples

Input

3
1 3 2

Output

1 3 2 

Input

4
2 2 3 3

Output

2 1 3 4 

Input

1
2

Output

1 

Note

In the first test the numeration is already a permutation, so there is no need to change anything.

In the second test there are two pairs of equal numbers, in each pair you need to replace one number.

In the third test you need to replace 2 by 1, as the numbering should start from one.

題目大意：

題意1：有n個物品需要標記，每個物品標號都不能相同，要從1開始，且保證充分利用之前的標記，輸出最後每個物品的標記。

題意2：給出一個數字n，接下來是n個數，把其中的重複的數或者大於n的數進行替換，使得整個數列是由1~n來組成的，可能會有多種答案，輸出其中任意一種。詳情直接看樣例。

解題報告：

    水題，打個標記處理一下就好了。

AC程式碼：

#include<bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 5;
int a[MAX];
bool vis[MAX];
int ans[MAX];
int main()
{
	int n;
	cin>>n;
	memset(ans,-1,sizeof(ans));
	memset(vis,0,sizeof(vis));
	for(int i = 1 ; i <= n ; i++){
		cin >> a[i];
		if(!vis[a[i]] && a[i] <= n){
			ans[i] = a[i];
			vis[a[i]]=1;
		}
	}
	int cur = 1;
	for(int i = 1 ; i <= n ; i ++){
		if(ans[i] == -1){
			for(int j = cur; j<=n ; j++){
				if(!vis[j]){
					cur = j;
					ans[i] =cur;
					vis[cur]=1;
					cur++;
					break;
				}
			}
		}
	}
	for(int i = 1; i<=n ; i++) {
		printf("%d%c",ans[i],i == n ? '\n' : ' ');
	}
		
	return 0;	
}