cf 569B. Inventory （模拟？）

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 nnumbers 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 

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.

给定n个元素，可能有重复，问最少变动多少个，把n个元素变成1--n的一个排列。

wa了一次。

需要注意的是，a[i]可能大于n

比如

4

5 6 7 8 

这样的数据也是存在的。

/*************************************************************************
    > File Name: code/cf/#315/B.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月11日 星期二 00时37分43秒
 ************************************************************************/
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=2E5+7;
int a[N];
int n;
bool vis[N];
int b[N];
int main()
{
    cin>>n;
    memset(vis,false,sizeof(vis));
    for ( int i = 0 ; i < n; i ++ )
    {
    scanf("%d",&a[i]);
    vis[a[i]]=true;
    }
    int cnt  = 0;
    for ( int i = 1 ; i <= n ; i++)
    {
    if (!vis[i])
    {
        cnt++;
        b[cnt] = i;
    }
    }
    memset(vis,false,sizeof(vis));
    cnt  = 1;
    for ( int i = 0 ; i < n ; i++)
    {
    if (vis[a[i]]||a[i]>n)
    {
        a[i] = b[cnt];
        cnt++;
    }
    else
    {
        vis[a[i]] = true;
    }
    }
    cout<<a[0];
    for ( int i = 1 ; i < n; i ++)
    {
    cout<<" "<<a[i];
    }
    cout<<endl;
    return 0;
}