C. Tanya and Toys_模拟

C. Tanya and Toys

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 7
1 3 4

output

2
2 5

解题报告：

1、数据达到10^9，开数组模拟是肯定不可以的。

#include <bits/stdc++.h>

using namespace std;

const int N=;
const int INF=;

map<int,bool> have;
vector<int> ans;

int main()
{
    int n,m,x;

    cin>>n>>m;

    for(int i=;i<=n;i++)
    {
        scanf("%d",&x);
        have[x]=true;
    }

    for(int i=;m>=i;i++)
    {
        if(have[i]) continue;
        ans.push_back(i);
        have[i]=true;
        m-=i;
    }

    cout<<ans.size()<<endl;
    for(int i=;i<ans.size();i++)
        printf("%d ",ans[i]);
    return ;
}