You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.

You want to find a subset of items to put such that:

* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.



Find the maximum total value of items in the knapsack.

Constraints

* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015

Input


N W
v1 w1
v2 w2
:
vN wN


The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.

Output

Print the maximum total values of the items in a line.

Examples

Input

4 5
4 2
5 2
2 1
8 3


Output

13


Input

2 20
5 9
4 10


Output

9
#include "bits/stdc++.h"

#define INF 200000000
#define FER 0.0000000001

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<double, double> pdd;

pll ps[1 << 20];
ll w[40], v[40];

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N;
	ll W;
	cin >> N >> W;
	
	for (int i = 0; i < N; i++)
	{
		cin >> v[i] >> w[i];
	}

	int n2 = N / 2;
	for (int i = 0; i < 1 << n2; i++)
	{
		ll sv = 0, sw = 0;
		for (int j = 0; j < n2; j++)
		{
			if (i >> j & 1)
			{
				sw += w[j];
				sv += v[j];
			}
		}
		ps[i] = make_pair(sw, sv);
	}

	sort(ps, ps + (1 << n2));
	int m = 1;
	for (int i = 1; i < 1 << n2; i++)
	{
		if (ps[m - 1].second < ps[i].second)
		{
			ps[m++] = ps[i];
		}
	}

	ll ans = 0;
	for (int i = 0; i < 1 << (N - n2); i++)
	{
		ll sw = 0, sv = 0;
		for (int j = 0; j < (N - n2); j++)
		{
			if (i >> j & 1)
			{
				sw += w[n2 + j];
				sv += v[n2 + j];
			}
		}
		if (sw <= W)
		{
			ll tv = (lower_bound(ps, ps + m, make_pair(W - sw, LLONG_MAX)) - 1)->second;
			ans = max(ans, sv + tv);
		}
	}

	cout << ans << endl;
	return 0;
}
