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                                      **B. Inventory**

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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.

Sample test(s)

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 

题目大意：就是有n个数。不能有比n大的数，然后尽可能的改变最少的步骤就能让它符合题意，注意a[i] >=1，当然不看也能够。

。

。。

解体思路：就是用两个数组，一个是标记的，另一个是将不符合的转化为符合的数。

详细详见代码：

/*

Date : 2015-8-20

Author : ITAK

Motto :

今日的我要超越昨日的我，明日的我要胜过今日的我；

以创作出更好的代码为目标，不断地超越自己。

*/

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

const int maxn = 1e5+5;

bool f[maxn];//推断是不是符合条件

int data[maxn];

int fac[maxn];//把不符合的改为符合的除0外

int main()

{

    int m;

    scanf("%d",&m);

    memset(f, 0, sizeof(f));

    int cnt = 1;//记录不符合的个数。从1開始啊。。。

for(int i=1; i<=m; i++)

    {

        scanf("%d",&data[i]);

        if(!f[data[i]] && data[i]<=m)

            f[data[i]] = 1;

        else

        {

            fac[cnt] = i;

            cnt++;

        }

    }

    cnt--;

    for(int i=m; i>0; i--)

    {

        if(!f[i])

        {

            data[fac[cnt]] = i;

            cnt--;

        }

    }

    for(int i=1; i<m; i++)

        printf("%d ",data[i]);

    printf("%d\n",data[m]);

    return 0;

}