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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

#define pb push_back
#define pf push_front
#define mp make_pair
#define fi(a, b) for(int i=a; i<=b; i++)
#define fj(a, b) for(int j=a; j<=b; j++)
#define fo(a, b) for(int o=a; o<=b; o++)
#define fdi(a, b) for(int i=a; i>=b; i--)
#define fdj(a, b) for(int j=a; j>=b; j--)
#define fdo(a, b) for(int o=a; o>=b; o--)
#define clr(x) memset(x, 0, sizeof(x))
#define cpy(x,y) memcpy(x, y, sizeof(y))
#define sz(x) (int)x.size()
#define all(x) (x).begin(), (x).end()

typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef vector<pii> vpii;
typedef vector<ll> vl;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;

char __buffer[100000];

#ifdef _DEBUG
#define LOCAL
#endif

void err(const char *format, ... ) {
#ifdef LOCAL
	va_list ap;
	va_start(ap, format);
	vsprintf(__buffer, format, ap);
	va_end(ap);
	
	fprintf(stderr, "\t%s", __buffer);
#else
	if (format < 0) {
		__buffer[0]++;
	}
#endif
}

void errln(const char *format, ... ) {
#ifdef LOCAL
	va_list ap;
	va_start(ap, format);
	vsprintf(__buffer, format, ap);
	va_end(ap);
	
	fprintf(stderr, "\t%s\n", __buffer);
#else
	if (format < 0) {
		__buffer[0]++;
	}
#endif
}

void errln() {
#ifdef LOCAL	
	fprintf(stderr, "\n");
#endif	
}

double START_TIME;

void exit() {
#ifdef LOCAL	
	cerr << "TIME: " << setprecision(5) << fixed << (clock() - START_TIME) / CLOCKS_PER_SEC << endl;
#endif	
	exit(0);
}

template<typename A, typename B>
ostream& operator<<(ostream& os, pair<A, B> p) {
	os << "(" << p.first << ", " << p.second << ")";
	return os;
}

template<typename T>
ostream& operator<<(ostream& os, vector<T> v) {
	fi(0, sz(v) - 1) {
		os << v[i] << " ";
	}
	return os;
}

template<typename T>
ostream& operator<<(ostream& os, set<T> t) {
	for (auto z : t) {
		os << z << " ";
	}
	return os;
}

template<typename T1, typename T2>
ostream& operator<<(ostream& os, map<T1, T2> t) {
	cerr << endl;
	for (auto z : t) {
		os << "\t" << z.first << " -> " << z.second << endl;
	}
	return os;
}

#ifdef LOCAL
#define dbg(x) {cerr << __LINE__ << "\t" << #x << ": " << x << endl;}
#define dbg0(x, n) {cerr << __LINE__ << "\t" << #x << ": "; for (int ABC = 0; ABC < n; ABC++) cerr << x[ABC] << ' '; cerr << endl;}
#define dbg1(x, n) {cerr << __LINE__ << "\t" << #x << ": "; for (int ABC = 1; ABC <= n; ABC++) cerr << x[ABC] << ' '; cerr << endl;}
#else
#define dbg(x) while(0){}
#define dbg0(x, n) while(0){}
#define dbg1(x, n) while(0){}
#endif

#ifdef LOCAL
#define ass(x) if (!(x)) { cerr << __LINE__ << "\tassertion failed: " << #x << endl, abort(); }
#else
#define ass(x) assert(x)
#endif

///////////////////////////////////////////////////

const int MAX = 200 * 1000 + 41;
const ll INF = 1000LL * 1000LL * 1000LL * 1000LL * 1000LL * 1000LL + 41;

int n;
int X;
ll ans;
int x[MAX];
ll pref[MAX];

ll mul(ll a, ll b) {
	if (a == 0 || b == 0) return 0;
	if (INF / a <= b) return INF;
	return a * b;
}

ll add(ll a, ll b) {
	return min(a + b, INF);
}

ll calc(int q) {
	ll res = (ll) X * (q + n);
	res = add(res, mul(5, (pref[n] - pref[n - q])));
	int p = n - q;
	int k = 1;
	while (p > 0) {
		int take = min(q, p);
		res = add(res, mul(pref[p] - pref[p - take], (2 * k + 3)));		
		k++;
		p -= take;
	}

	return res;
}

void solve() {	
	ans = INF;
	fi(1, n) {
		ans = min(ans, calc(i));
	}
	printf("%lld\n", ans);
}

int main() {
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	START_TIME = (double)clock();
#endif

  scanf("%d %d", &n, &X);
	fi(1, n) {
		scanf("%d", &x[i]);
	}
	fi(1, n) {
		pref[i] = pref[i - 1] + x[i];
	}
	solve();

	exit();
	return 0;
}