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 Python (3.4.3) (judge verdict: Accepted)

import sys
import math
import collections
import itertools
import array
import inspect

# Set max recursion limit
sys.setrecursionlimit(1000000)


# Debug output
def chkprint(*args):
    names = {
        id(v): k
        for k, v in inspect.currentframe().f_back.f_locals.items()
    }
    print(', '.join(
        names.get(id(arg), '???') + ' = ' + repr(arg) for arg in args))


# Binary converter
def to_bin(x):
    return bin(x)[2:]


def li_input():
    return [int(_) for _ in input().split()]


def gcd(n, m):
    if n % m == 0:
        return m
    else:
        return gcd(m, n % m)


def gcd_list(L):
    v = L[0]

    for i in range(1, len(L)):
        v = gcd(v, L[i])

    return v


def lcm(n, m):
    return (n * m) // gcd(n, m)


def lcm_list(L):
    v = L[0]

    for i in range(1, len(L)):
        v = lcm(v, L[i])

    return v


# Width First Search (+ Distance)
def wfs_d(D, N, K):
    """
    D: 隣接行列(距離付き)
    N: ノード数
    K: 始点ノード
    """

    dfk = [-1] * (N + 1)
    dfk[K] = 0

    cps = [(K, 0)]
    r = [False] * (N + 1)
    r[K] = True
    while len(cps) != 0:
        n_cps = []
        for cp, cd in cps:
            for i, dfcp in enumerate(D[cp]):
                if dfcp != -1 and not r[i]:
                    dfk[i] = cd + dfcp
                    n_cps.append((i, cd + dfcp))
                    r[i] = True

        cps = n_cps[:]

    return dfk


# Depth First Search of Tree (+Distance)
def dfs_d(v, pre, dist):
    """
    v:  現在のノード
    pre: １つ前のノード
    dist: 現在の距離

    以下は別途用意する
    D: 隣接リスト(行列ではない)
    D_dfs_d: dfs_d関数で用いる，始点ノードから見た距離リスト
    """

    global D
    global D_dfs_d

    D_dfs_d[v] = dist

    for next_v, d in D[v]:
        if next_v != pre:
            dfs_d(next_v, v, dist + d)

    return


def sigma(N):
    ans = 0
    for i in range(1, N + 1):
        ans += i
    return ans


def comb(n, r):
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n

    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]

    for p in range(2, r + 1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p - 1, r, p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot

    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])

    return result


# --------------------------------------------

dp = None


def main():
    N, x = li_input()
    A = sorted(li_input())

    ans = 0

    for i, a in enumerate(A):
        if x >= a:
            if i != N - 1 or i == N-1 and x == a:
                x -= a
                ans += 1

    print(ans)


main()