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>
using namespace std;
using ll = long long;

const int MAX_N = 40;

ll w[MAX_N],v[MAX_N];

int n;
ll W;

pair<ll,ll> ps[1<<(MAX_N/2)];

void solve()
{
	int n2 = n/2;
	for(int i = 0;i<(1<<n2);i++)
	{
		ll sw = 0, sv = 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 res = 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;
			res = max(res,sv+tv);
		}
	}
	cout<<res<<endl;
}

int main()
{
	cin >> n >> W;
	for(int i = 0;i<n;i++)
	{
		cin >> v[i] >> w[i];
	}
	solve();
}
