A. Amr and Music

解题思路：给出n种乐器学习所需要的时间，以及总共的天数， 问最多能够学多少门乐器，并且输出这几门乐器在原序列中的序号(不唯一)

按照升序排序，为了学到最多的乐器，肯定要选择花费时间最少的来学习 然后用结构体来储存该门乐器在原序列中的序号。

A. Amr and Music

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.

Amr has n instruments, it takes a_i days to learn i-th instrument. Being busy, Amr dedicated k days to learn how to play the maximum possible number of instruments.

Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.

Input

The first line contains two numbers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.

The second line contains n integers a_i (1 ≤ a_i ≤ 100), representing number of days required to learn the i-th instrument.

Output

In the first line output one integer m representing the maximum number of instruments Amr can learn.

In the second line output m space-separated integers: the indices of instruments to be learnt. You may output indices in any order.

if there are multiple optimal solutions output any. It is not necessary to use all days for studying.

Sample test(s)

input

4 10 4 3 1 2

output

4 1 2 3 4

input

5 6 4 3 1 1 2

output

3 1 3 4

input

1 3 4

output

0

Note

In the first test Amr can learn all 4 instruments.

In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.

In the third test Amr doesn't have enough time to learn the only presented instrument.

/*287 div2 A*/
#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<algorithm>  
using namespace std;
struct Node
{
	int first;
	int second;
} a[105];
bool cmp(Node n1,Node n2){  
    return n1.first<n2.first;  
}  

int main()
{
	int n,k,i,j,sum=0;
	scanf("%d %d",&n,&k);
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i].first);
			a[i].second=i;		
		}
		sort(a+1,a+n+1,cmp);
		for(i=1;i<=n;i++)
		{
			if(sum+a[i].first<=k)
			sum=sum+a[i].first;
			else
			break;
		}
		printf("%d
",i-1);
		for(j=1;j<i;j++)
		printf("%d ",a[j].second);	
}