Problem:
Score : 200 points

Problem Statement

There are N children, numbered 1, 2, ..., N.

Snuke has decided to distribute x sweets among them.
He needs to give out all the x sweets, but some of the children may get zero sweets.

For each i (1 \leq i \leq N), Child i will be happy if he/she gets exactly a_i sweets.
Snuke is trying to maximize the number of happy children by optimally distributing the sweets.
Find the maximum possible number of happy children.

Constraints

All values in input are integers.

2 \leq N \leq 100

1 \leq x \leq 10^9

1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N x
a_1 a_2 ... a_N

Output

Print the maximum possible number of happy children.

Sample Input 1

3 70
20 30 10

Sample Output 1

2

One optimal way to distribute sweets is (20, 30, 20).

Sample Input 2

3 10
20 30 10

Sample Output 2

1

The optimal way to distribute sweets is (0, 0, 10).

Sample Input 3

4 1111
1 10 100 1000

Sample Output 3

4

The optimal way to distribute sweets is (1, 10, 100, 1000).

Sample Input 4

2 10
20 20

Sample Output 4

0

No children will be happy, no matter how the sweets are distributed.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
    問題をよく読もう！
    論理的に考えよう！
    サンプルを確認しよう!
    絶対に諦めるな！
    工夫をしろ！
    配列は少し多めにとっておく

    Twitterは終わるまでログアウト！
    （間違えて解法をツイートしてはいけないから）

*/
//include
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
#include<stdio.h>
#include<cstring>
//namespace
using namespace std;
//繰り返し
#define REP(i, m, n) for(int i=(int)m; i<(int)n; ++i)
#define rep(i, n) REP(i, 0, n)
//イテレータ
#define all_range(C) begin(C), end(C)
//簡略化
typedef long long ll;
typedef pair<int,int> pint;
typedef pair<ll,int> pli;
typedef pair<string,int> pst;

const int inf = 1e9+7;
const ll longinf = 1LL<<60;
const ll mod = 1e9+7;
//最大最小
template<typename T1, typename T2> inline void chmin(T1 &a, T2 b) {if(a>b) a=b;}
template<typename T1, typename T2> inline void chmax(T1 &a, T2 b) {if(a<b) a=b;}

//関数

//最大公約数を求める
ll my_gcd(ll x, ll y) {
  ll r;

  while(y!=0) {
    r = x%y;
    x = y;
    y = r;
  }

  return x;
}

//最小公約数を求める
ll my_lcm(ll x, ll y) {

  return (x*y)/my_gcd(x,y);
}

//xのy乗を求める ※参考：10^18乗まで
ll my_pow(ll x, ll y) {
  ll sum = 1;

  while(y>0) {
    sum *= x;
    --y;
  }

  return sum;
}

/*考えを書くスペース
  累乗和で解く

  ・総和を求める
  ・取り消すところの一つ先を見る
*/
//main.cpp----------------------------
#define max_n 100

void solve();

ll N, x;
ll a[max_n+1];

int main() {
  cin >> N >> x;
  rep(i, N) cin >> a[i];

  solve();

  return 0;
}

void solve() {

  sort(a, a+N);

  int i = 0, cnt = 0;
  while(i<N-1&&x>=a[i]) {
    x -= a[i];
    ++cnt;
    ++i;
  }

  if(x==a[N-1]) ++cnt;

  cout << cnt << endl;
}