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 <algorithm>
#include <cstdio>
typedef long long ll;
using namespace std;
static const int MAX = 41;
static const long long INFTY = 100000000000000001;
int n;
ll w[MAX], v[MAX];
ll W;
pair <ll, ll> ps[1 << (MAX/2)];

int main()
{
    cin >> n >> W;
    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, 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];
    }

    long long 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, INFTY)) - 1) -> second;
            res = max(res, sv + tv);
        }
    }
    cout << res << endl;
    return 0;
}

