Codeforces Round #346 (Div. 2) C.Tanya and Toys

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In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to109. A toy from the new collection of the i-th type costs i bourles.

Tania has managed to collect n different types of toys a1, a2, ..., an from the new collection. Today is Tanya's birthday, and her mother decided to spend no more than m bourles on the gift to the daughter. Tanya will choose several different types of toys from the new collection as a gift. Of course, she does not want to get a type of toy which she already has.

Tanya wants to have as many distinct types of toys in her collection as possible as the result. The new collection is too diverse, and Tanya is too little, so she asks you to help her in this.

Input

The first line contains two integers n (1 ≤ n ≤ 100 000) and m (1 ≤ m ≤ 109) — the number of types of toys that Tanya already has and the number of bourles that her mom is willing to spend on buying new toys.

The next line contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the types of toys that Tanya already has.

Output

In the first line print a single integer k — the number of different types of toys that Tanya should choose so that the number of different types of toys in her collection is maximum possible. Of course, the total cost of the selected toys should not exceed m.

In the second line print k distinct space-separated integers t1, t2, ..., tk (1 ≤ ti ≤ 109) — the types of toys that Tanya should choose.

If there are multiple answers, you may print any of them. Values of ti can be printed in any order.

Examples
input
3 71 3 4
output
22 5 
input
4 144 6 12 8
output
47 2 3 1

Note

In the first sample mom should buy two toys: one toy of the 2-nd type and one toy of the 5-th type. At any other purchase for 7 bourles (assuming that the toys of types 1, 3 and 4 have already been bought), it is impossible to buy two and more toys.

题意:总共有1-10^9中玩具,每种玩具的价格便是他的序号,现在给出n,m, (1 ≤ n ≤ 100000) and m (1 ≤ m ≤ 10^9),接下来n个数,表示现在他已经拥有的玩具的序号,然后给你m元,问最后还能买多少个玩具(不能买已经拥有的)

思路:

方法一:由于玩具的序号比较大,我们不能用数组来存下,而一开始拥有的玩具数又较少,所以我们可以用map来表示一开始他拥有哪些玩具

又因为他的钱只有10^9,所以买的玩具的最大数目一定不会超过10^6,(因为1+2+...+10^6远远大于10^9),所以只需要暴力枚举1-10^6的玩具,直到钱用完。

时间复杂度为10^6*log n

方法二:

对原来给出的玩具进行排序,贪心取即可(即如果现在的i和a[j]相等,i++,j++),如果不相等,就表示可以买i。

方法一的代码

#include<bits/stdc++.h>using namespace std;const int maxn=1100000;map<int,int>mp;int ans[maxn];int main(){    int n,m,x;    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++){        scanf("%d",&x);        mp[x]=1;    }    int tot=0;    for(int i=1;i;i++){        if(mp[i]==1)            continue;        if(m<i)            break;        m-=i;        ans[++tot]=i;    }    printf("%d\n",tot);    for(int i=1;i<=tot;i++){        if(i==tot)            printf("%d\n",ans[tot]);        else            printf("%d ",ans[i]);    }}

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