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<bits/stdc++.h>

using namespace std;

typedef long long int64;

int N;
int64 v[40], w[40], W, ret;

void dfs(int idx, int sz, int64 vv, int64 ww, vector< pair< int64, int64 > > &kk)
{
  if(idx == sz) {
    if(ww <= W) {
      kk.emplace_back(ww, vv);
      ret = max(ret, vv);
    }
    return;
  }
  dfs(idx + 1, sz, vv + v[idx], ww + w[idx], kk);
  dfs(idx + 1, sz, vv, ww, kk);
}

int main()
{
  cin >> N >> W;
  for(int i = 0; i < N; i++) cin >> v[i] >> w[i];

  vector< pair< int64, int64 > > a, b;
  dfs(0, N >> 1, 0, 0, a);
  dfs(N >> 1, N, 0, 0, b);
  sort(begin(b), end(b));

  for(int i = 1; i < b.size(); i++) {
    b[i].second = max(b[i].second, b[i - 1].second);
  }
  
  for(auto &ss : a) {
    auto get = lower_bound(begin(b), end(b), make_pair(W - ss.first + 1, 0LL));
    if(get != begin(b)) {
      --get;
      if(ss.first + get->first > W) continue;
      ret = max(ret, ss.second + get->second);
    }
  }
  cout << ret << endl;
}