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 <vector>
#include <map>

using namespace std;

long long int v[45], w[45];

int main()
{
    int N;
    long long int W;
    cin >> N >> W;
    for(int i = 0; i < N; ++i){
        cin >> v[i] >> w[i];
    }
    map<long long int, long long int> m1, m2;
    int half = N / 2;
    for(int i = 0; i < (1 << half); ++i){
        long long int nv = 0, nw = 0;
        int now = i;
        for(int j = 0; now; j++){
            if(now % 2){
                nv += v[j];
                nw += w[j];
            }
            now /= 2;
        }
        if(nw > W) continue;
        m1[nw] = max(m1[nw], nv);
    }
    for(int i = 0; i < (1 << (N - half)); ++i){
        long long int nv = 0, nw = 0;
        int now = i;
        for(int j = half; now; j++){
            if(now % 2){
                nv += v[j];
                nw += w[j];
            }
            now /= 2;
        }
        if(nw > W) continue;
        m2[nw] = max(m2[nw], nv);
    }
    long long int mx;
    mx = 0;
    for(auto& p : m1){
        mx = max(mx, p.second);
        p.second = mx;
    }
    mx = 0;
    for(auto& p : m2){
        mx = max(mx, p.second);
        p.second = mx;
    }
    long long int ans = 0;
    for(auto& p : m1){
        if(p.first <= W){
            auto it = m2.upper_bound(W - p.first);
            it = prev(it);
            if(m2.find(W - p.first) != m2.end()){
                it = m2.find(W - p.first);
            }
            //cout << W - p.first << " " << it->first << endl;
            ans = max(ans, p.second + it->second);
        }
    }
    cout << ans << endl;
}