Codeforces Round #315 (Div. 2)【贪心/重排去掉大于n的元素和替换重复的元素】

B. Inventory

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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, and no two numbers are equal.

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.

【题意】：对于给定一系列有编号的物品，他们的编号可能重复，也有可以超过了物品总个数，现在要求给所有物品重新编号，让所有物品编号都在个数范围内，并且不重复，修改次数尽可能小，输出修改后的编号。

【分析】：大于的去掉，重复的替换

【代码】：

#include <bits/stdc++.h>

using namespace std;
const int N = 1e5 + 100;
int n;
int a[N];
bool vis[N];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if( !vis[a[i]] && a[i]<=n )
            vis[a[i]]=1;
        else
            a[i] = -1;
    }
    int cnt=1;
    for(int i=1;i<=n;i++)
    {
        if(a[i]==-1){
        while(vis[cnt]) cnt++;
            a[i]=cnt;
        cnt++;
        }
    }
    for(int i=1;i<=n-1;i++)
        printf("%d ",a[i]);
    printf("%d
",a[n]);
    return 0;
}