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 ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back
#define MP make_pair
#define MOD 1000000007
#define INF (1LL<<30)
#define LLINF (1LL<<60)
#define PI 3.14159265359
#define EPS 1e-12
#define int ll

int v[45], w[45];
pair<ll, ll> ps[1<<20];
signed main(void)
{
  int n, W;
  cin >> n >> W;
  REP(i, n) cin >> v[i] >> w[i];

  int n2 = n/2;
  REP(i, 1<<n2) {
    int sw = 0, sv = 0;
    REP(j, n2) {
      if(i>>j&1) {
        sw += w[j];
        sv += v[j];
      }
    }
    ps[i] = MP(sw, sv);
  }
  sort(ps, ps+(1<<n2));
  int m=1;
  REP(i, 1<<n2) {
    if(ps[m-1].second < ps[i].second) {
      ps[m++] = ps[i];
    }
  }
  int res = 0;
  REP(i, 1<<(n-n2)) {
    int sw = 0, sv = 0;
    REP(j, n-n2) {
      if(i>>j&1) {
        sw += w[n2+j];
        sv += v[n2+j];
      }
    }
    if(sw <= W) {
      int tv = (lower_bound(ps, ps+m, MP(W-sw, INF))-1)->second;
      res = max(res, sv+tv);
    }
  }
  cout << res << endl;
  return 0;
}