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<math.h>
#include<vector>

using namespace std;

#define INF 40000000000000007

/*use prepaired combination*/
vector< pair<long long, long long > > A,B;

long long v[40][2];
int n;
long long w;

int main(){
  cin >> n >> w;
  for(int i = 0; i < n ; i++){
    cin >> v[i][1] >> v[i][0];//ここ注意。これで重さ順のソートが楽になる。
  }
  int a = min(20, n);
  int b = max(n-20, 0);
  A.push_back(make_pair(0,0));
  B.push_back(make_pair(0,0));
  for(int i = 0; i < a; i++){
    for(int j = 0; j < pow(2, i); j++){
      A.push_back(make_pair(A[j].first+v[i][0], A[j].second+v[i][1]));
    }
  }
  for(int i = 0; i < b; i++){
    for(int j = 0; j < pow(2, i); j++){
      B.push_back(make_pair(B[j].first+v[i+a][0], B[j].second+v[i+a][1]));
    }
  }
  sort(A.begin(), A.end());
  long long res = 0;
  int m = 1;
  for(int i = 1; i < 1 << a; i++){
    if(A[m-1].second < A[i].second){
      A[m++] = A[i];
    }
  }

  for(int i = 0; i < B.size(); i++){
    if(B[i].first <= w){
      const pair<long long, long long> r = make_pair(w - B[i].first, INF);
      auto tv = (lower_bound(A.begin() , A.begin()+m, r) - 1);
      long long t = tv->second;
      res = max(res, B[i].second + t);
    }
  }

  cout << res << endl;
}

