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 <iostream>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <map>
#include <stack>
#include <queue>
#include <set>
#include <cstring>

using namespace std;
// ascending order
#define vsort(v) sort(v.begin(), v.end())
// descending order
#define vsort_r(v) sort(v.begin(), v.end(), greater<int>())
#define vunique(v) unique(v.begin(), v.end())
#define mp make_pair
#define ts(x) to_string(x)
#define rep(i, a, b) for(int i = (int)a; i < (int)b; i++)
#define repm(i, a, b) for(int i = (int)a; i > (int)b; i--)
#define bit(a) bitset<8>(a)
#define des_priority_queue priority_queue<int, vector<int>, greater<int> >
typedef long long ll;
typedef pair<int, int> P;
const ll INF = 1e18;

#define MAX_V 1000000

#define MAX_N 40
int n;
ll w[MAX_N], v[MAX_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, INF)) - 1)->second;
			res = max(res, sv + tv);
		}
	}

	cout << res << endl;
}

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

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

