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<algorithm>
using namespace std;
typedef long long ll;

int main(){
    int n;
    ll W;
    cin >> n >> W;
    vector<ll> v(n), w(n);
    for(int i = 0; i < n; i++){
        cin >> v[i] >> w[i];
    }
    int n2 = n/2;
    vector<pair<ll,ll>> vp;
    for(int i = 0; i < 1<<n2; i++){
        ll tmpv = 0, tmpw = 0;
        for(int j = 0; j < n2; j++){
            if((i>>j)&1)    tmpv += v[j], tmpw += w[j];
        }
        if(tmpw > W)    continue;
        vp.push_back({tmpw, tmpv});
    }
    sort(vp.begin(), vp.end());
    for(int i = vp.size()-1; i >= 1; i--){
        if(vp[i].first == vp[i-1].first){
            int rm = i-1;
            while(vp[rm].first == vp[i].first)  vp[rm--] = {0,0};
            i = rm+1;
        }
    }
    sort(vp.begin(), vp.end());
    vp.erase(unique(vp.begin(),vp.end()), vp.end());
    
    for(int i = 1; i < vp.size(); i++){
        vp[i].second = max(vp[i].second, vp[i-1].second);
    }
    ll ans = 0;
    for(int i = 0; i < 1<<(n-n2); i++){
        ll tmpv = 0, tmpw = 0;
        for(int j = 0; j < n-n2; j++){
            if((i>>j)&1)    tmpv += v[n2+j], tmpw += w[n2+j];
        }
        if(tmpw > W)    continue;
        auto it = upper_bound(vp.begin(),vp.end(),make_pair(W-tmpw+1,0ll));
        it--;
        ans = max(ans, (*it).second+tmpv);
    }
    cout << ans << endl;
    return 0;
}
