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 999999999999999
#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>pll;

const int MAX_N = 40;

ll w[MAX_N];
ll v[MAX_N];
ll W;
int n;
pll ps[1<<(MAX_N/2)];

void solve()
{
	int n2 = n/2;
	rep(i,1<<n2){
		ll sw =0,sv=0;
		rep(j,n2){
			if(i&(1<<j)){
				sw += w[j];
				sv += v[j];
			}
		}
		ps[i] = pll(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;
	rep(i,1<<(n-n2)){
		ll sw=0,sv=0;
		rep(j,n-n2){
			if((i>>j)&1){
				sw += w[n2+j];
				sv += v[n2+j];
			}
		}
		if(sw <= W){
			ll tv = (lower_bound(ps,ps+m,pll(W-sw,INF))-1)->second;
			res = max(res,sv+tv);
		}
	}
	printf("%lld\n",res);
}

int main()
{
	cin >> n >> W;
	rep(i,n){
		cin >> v[i] >> w[i];
	}
	solve();
	return 0;
}