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 ll long long
#define INF 999999999
#define MOD 1000000007
#define rep(i,n) for(int i=0;i<n;i++)

using namespace std;

typedef pair<int,int>P;
typedef pair<ll,ll>pl;

const int MAX_N = 40;

ll v[MAX_N];
ll w[MAX_N];
vector<pl> x,y;

void solve(int s,int t,vector<pl> &x)
{
	int z = t-s;
	rep(i,1<<z){
		pl p = pl(0,0);
		rep(j,z){
			if(i&(1<<j)){
				p.first += w[s+j];
				p.second += v[s+j];
			}
		}
		x.push_back(p);
	}
}

int main()
{
	int n;
	ll lm;
	cin >> n >> lm;
	rep(i,n){
		cin >> v[i] >> w[i];
	}
	solve(0,n/2,x);
	solve(n/2,n,y);
	sort(x.begin(),x.end());
	x.erase(unique(x.begin(),x.end()),x.end());
	sort(y.begin(),y.end());
	ll t=0;
	ll maxv = -(1LL<<60);
	ll ans = 0;
	for(int i=x.size()-1;i>=0;i--){
		while(t < y.size() && x[i].first+y[t].first <= lm){
			maxv = max(maxv,y[t].second);
			t++;
		}
		ans = max(ans,maxv+x[i].second);
	}
	cout << ans << endl;
	return 0;
}