Tanya and Toys （贪心）

http://codeforces.com/problemset/problem/659/C

In Berland recently a new collection of toys went on sale. This collection consists of 109 types of toys, numbered with integers from 1 to 109. 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 

Input

4 14
4 6 12 8

Output

4
7 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.

贪心是一种思想，即在对问题求解时，总是做出在当前看来是最好的选择。也就是说，不从整体最优上加以考虑，他所做出的是在某种意义上的局部最优解。

i从1开始选最小的，让当前所拥有的钱与i比较，i与a[ j ]比较时，并不需要遍历a[ ]中所有的元素，把a[ ]中的元素从小到大排列，i也是从小到大的，所以当i=a[ j ],令j++即可，两层遍历会造成超时，诶，以后一定要少用多层循环，一道水题用了大半个下午，革命尚未成功，还需要多多努力呀！

#include <iostream>
#include<Cstring>
#include<Cstdio>
#include<algorithm>

using namespace std;

int a[1000050];
int b[1000050];
int main()
{
    int n,m;
    int i,j,k=0;
    int sum=0;
    int c=1;

    scanf("%d%d",&n,&m);
    for(i=0;i<n;++i)
    {
        scanf("%d",&a[i]);
    }
    sort(a,a+n);
    for(i=1,j=0;;++i)
    {
        if(j>=n) j=n-1;
        if(i==a[j]) {j++;continue;}
        else  if(m<i) break;
              else m=m-i,b[k++]=i;
    }

    cout<<k<<endl;
    for(i=0;i<k;++i)
    {
        printf("%d ",b[i]);
    }
    printf("\n");

    return 0;
}