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 <vector>
#include <map>
#include<algorithm>
#include <climits>
typedef long long ll;
#define INF (LLONG_MAX/10)
using namespace std;

int main() {
	cin.tie(0); ios::sync_with_stdio(false);
	ll N, W; cin >> N >> W;
	vector<ll> w(N + 1), v(N + 1);
	vector<pair<ll, ll>> ps(1 << ((N+1)/ 2));
	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 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.begin(), ps.begin() + (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.begin(), ps.begin() + m, make_pair(W - sw, INF)) - 1)->second;
			res = max(res, sv + tv);
		}
	}
	cout << res << endl;
}