題干：

Ivan has an array consisting of?n?elements. Each of the elements is an integer from?1?to?n.

Recently Ivan learned about permutations and their lexicographical order. Now he wants to change (replace)?minimum?number of elements in his array in such a way that his array becomes a?permutation?(i.e. each of the integers from?1?to?n?was encountered in his array exactly once). If there are multiple ways to do it he wants to find the?lexicographically minimal?permutation among them.

Thus minimizing the number of changes has the first priority, lexicographical minimizing has the second priority.

In order to determine which of the two permutations is lexicographically smaller, we compare their first elements. If they are equal — compare the second, and so on. If we have two permutations?x?and?y, then?x?is lexicographically smaller if?xi?<?yi, where?i?is the first index in which the permutations?x?and?y?differ.

Determine the array Ivan will obtain after performing all the changes.

Input

The first line contains an single integer?n?(2?≤?n?≤?200?000) — the number of elements in Ivan's array.

The second line contains a sequence of integers?a1,?a2,?...,?an?(1?≤?ai?≤?n) — the description of Ivan's array.

Output

In the first line print?q?— the minimum number of elements that need to be changed in Ivan's array in order to make his array a permutation. In the second line, print the?lexicographically minimal?permutation which can be obtained from array with?qchanges.

Examples

Input

4 3 2 2 3

Output

2 1 2 4 3

Input

6 4 5 6 3 2 1

Output

0 4 5 6 3 2 1

Input

10 6 8 4 6 7 1 6 3 4 5

Output

3 2 8 4 6 7 1 9 3 10 5

Note

In the first example Ivan needs to replace number three in position?1?with number one, and number two in position?3?with number four. Then he will get a permutation?[1, 2, 4, 3]?with only two changed numbers — this permutation is lexicographically minimal among all suitable.

In the second example Ivan does not need to change anything because his array already is a permutation.

題目大意：

給出n的數(shù)的序列，這n個(gè)數(shù)范圍為1~n。 現(xiàn)在問最少改變幾個(gè)數(shù)能使這n個(gè)數(shù)成為1~n的某個(gè)全排列，若有多種情況，要求使全排列的字典序最小。輸出改變的數(shù)的個(gè)數(shù)和這個(gè)全排列。

解題報(bào)告：

涉及字典序最小的問題其實(shí)比較好構(gòu)造，就是優(yōu)先貪心前面的，因?yàn)榍懊娴淖冃×艘獌?yōu)秀于后面無論多花里胡哨的操作。

首先需要改變的數(shù)很好求出，然后對(duì)于輸出全排列，就先把需要添加的數(shù)求出來，排個(gè)序，然后從前到后對(duì)原序列貪心搞一搞就好了。vis數(shù)組考慮這個(gè)數(shù)是否出現(xiàn)過。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define FF first #define SS second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; int n,a[MAX],cnt[MAX],res[MAX],tot,ans,vis[MAX]; int main() {cin>>n;for(int i = 1; i<=n; i++) scanf("%d",a+i),cnt[a[i]]++;for(int i = 1; i<=n; i++) {if(cnt[i] == 0) res[++tot] = i;if(cnt[i] > 1) ans += cnt[i]-1;}sort(res+1,res+tot+1);int cur = 1;for(int i = 1; i<=n; i++) {if(cnt[a[i]] == 1) continue;if(vis[a[i]]) cnt[a[i]]--,a[i] = res[cur++];else {if(res[cur] < a[i]) cnt[a[i]]--,a[i] = res[cur++];else vis[a[i]]=1;}}printf("%d\n",ans);for(int i = 1; i<=n; i++) printf("%d ",a[i]);return 0 ; }

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