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 <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
#include <cstring>
#include <deque>
#include <iomanip>
#include <limits>
#include <fstream>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
#define CLR(mat) memset(mat, 0, sizeof(mat))
typedef long long ll;
typedef pair<ll, ll> P; // 重さ、価値
vector<P> vec;
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll N, W;
  cin >> N >> W;
  ll v[N], w[N];
  FOR(i,0,N) cin >> v[i] >> w[i];
  int N2 = N / 2;
  FOR(i,0,1<<N2) {
    ll sum_v = 0, sum_w = 0;
    FOR(j,0,N2) {
      if(i&(1<<j)) {
        sum_v += v[j];
        sum_w += w[j];
      }
    }
    vec.push_back(P(sum_w, sum_v));
    //cout << sum_w << " " << sum_v << endl;
  }
  sort(vec.begin(), vec.end());
  ll max_v = 0;
  FOR(i,0,vec.size()) {
    if(vec[i].second < max_v) {
      vec[i].second = max_v;
    } else {
      max_v = vec[i].second;
    }
    // cout << vec[i].first << " " << max_v << endl;
  }
  ll ans = 0;
  FOR(i,0,1<<(N-N2)) {
    ll sum_v = 0, sum_w = 0;
    FOR(j,0,N-N2) {
      if(i&(1<<j)) {
        sum_v += v[j+N2];
        sum_w += w[j+N2];
      }
    }
    int k = upper_bound(vec.begin(), vec.end(), P(W - sum_w, 1e17)) - vec.begin();
    //cout << "upper " << W - sum_w << " " << sum_v << " " << k << endl;
    k = max(k - 1, 0);
    if(sum_w + vec[k].first <= W) {
      ans = max(ans, sum_v + vec[k].second);
    }
  }
  cout << ans << endl;
}
