題干：

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.

題目大意：

題意1：有n個(gè)物品需要標(biāo)記，每個(gè)物品標(biāo)號(hào)都不能相同，要從1開始，且保證充分利用之前的標(biāo)記，輸出最后每個(gè)物品的標(biāo)記。

題意2：給出一個(gè)數(shù)字n，接下來(lái)是n個(gè)數(shù)，把其中的重復(fù)的數(shù)或者大于n的數(shù)進(jìn)行替換，使得整個(gè)數(shù)列是由1~n來(lái)組成的，可能會(huì)有多種答案，輸出其中任意一種。詳情直接看樣例。

解題報(bào)告：

? ? 水題，打個(gè)標(biāo)記處理一下就好了。

AC代碼：

#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; }

?

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