CodeForces - 569B.Inventory

思路：这道题要重复数字的下标存起来，把没出现过的数字标记起来并且存起来，然后再把重复的数字换成没出现过的（没出现过的数字的数量 = 重复数字的数量）

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, ..., a_n (1 ≤ a_i ≤ 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.

#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <queue>
using namespace std;
int a[100000+8], sign[100000+8], tag[100000+8], position[100000+8];
int main()
{
    int n, miao = 0, root = 0;
    memset(sign, 0, sizeof(sign));
    scanf("%d", &n);
    for(int i = 0; i<n; i++)
    {
        scanf("%d", &a[i]);
        if(!sign[a[i]] && a[i] <= n && a[i]) sign[a[i]] = 1;
        else tag[miao++] = i;//把重复出现的数字的下标存起来
    }
    for(int i = 1; i <= n; i++)
        if(!sign[i]) position[root++] = i;//把没出现过的数字存起来
    for(int i = 0; i<miao; i++)//重复数字的个数与没出现过的数字的个数相等
    {
        a[tag[i]] = position[i];//把重复出现的数字的位置换上没出现过的最小的数字
    }
    int flag = 0;//输出
    for(int i = 0; i<n; i++)
    {
        if(flag)printf(" ");
        flag = 1;
        printf("%d", a[i]);
    }
    printf("\n");
    return 0;
}