CodeForces-659C-Tanya and Toys

Description

In Berland recently a new collection of toys went on sale. This collection consists of10⁹ types of toys, numbered with integers from1 to 10⁹. A toy from the new collection of thei-th type costs i bourles.

Tania has managed to collect n different types of toysa₁, a₂, ..., a_n 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) andm (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 integersa₁, a₂, ..., a_n (1 ≤ a_i ≤ 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 integerst₁, t₂, ..., t_k (1 ≤ t_i ≤ 10⁹) — the types of toys that Tanya should choose.

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

Sample Input

Input

3 71 3 4

Output

22 5 

Input

4 144 6 12 8

Output

47 2 3 1

有n种已经拥有的物品和m元钱，第i种物品价格为i。问你最多可以拥有的物品种类。其实也就是问没有选过的toy能最多选多少个。

很简单的贪心，从最小的找起就好了，但是1e9有点大，数组开不下，用了vector

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>using namespace std;vector <int> ans;bool vis[1111111];int main(){    int n,m,num;    while(scanf("%d%d",&n,&m)!=EOF)    {        memset(vis,0,sizeof(vis));        ans.clear();        for(int i=0;i<n;i++)        {            scanf("%d",&num);            if (num <= 1000000)            vis[num]=1;        }        for(int i=1;m-i>=0;i++)        {            if(!vis[i])            {                m-=i;            ans.push_back (i);            }        }        printf ("%d\n", ans.size ());        for (int i = 0; i < ans.size (); i++) {            printf ("%d%c", ans[i], i == (ans.size ()-1) ? '\n':' ');    }    }    return 0;}