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<cstdio>
#include<utility>
#include<algorithm>
using namespace std;
typedef long long ll;
static const int MAX_N = 40;
static const ll INF = 1 << 30;

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);
		}
	}

	printf("%lld\n", res);
}

int main(){
	scanf("%d %lld", &n, &W);
	for(int i = 0; i < n; i++){
		scanf("%lld %lld", &v[i], &w[i]);
	}
	solve();
	return 0;
}