Problem:
Score : 400 points

Problem Statement

You have N items and a bag of strength W.
The i-th item has a weight of w_i and a value of v_i.

You will select some of the items and put them in the bag.
Here, the total weight of the selected items needs to be at most W.

Your objective is to maximize the total value of the selected items.

Constraints

1 ≤ N ≤ 100

1 ≤ W ≤ 10^9

1 ≤ w_i ≤ 10^9

For each i = 2,3,...,N, w_1 ≤ w_i ≤ w_1 + 3.

1 ≤ v_i ≤ 10^7

W, each w_i and v_i are integers.

Input

Input is given from Standard Input in the following format:

N W
w_1 v_1
w_2 v_2
:
w_N v_N

Output

Print the maximum possible total value of the selected items.

Sample Input 1

4 6
2 1
3 4
4 10
3 4

Sample Output 1

11

The first and third items should be selected.

Sample Input 2

4 6
2 1
3 7
4 10
3 6

Sample Output 2

13

The second and fourth items should be selected.

Sample Input 3

4 10
1 100
1 100
1 100
1 100

Sample Output 3

400

You can take everything.

Sample Input 4

4 1
10 100
10 100
10 100
10 100

Sample Output 4

0

You can take nothing.

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <iostream>

using namespace std;

const long MAX_N = 100; // nの最大値

// 入力
long n, W;
long w[MAX_N], v[MAX_N];

// i番目以降の品物から重さの和がj以下なるように選んだときの、
// 取りうる価値の総和の最大値を返す関数
long rec(long i, long j) {
  long res;
  if (i == n) {
    // 品物がもう残っていないときは、価値の和の最大値は0で確定
    res = 0;
  } else if (j < w[i]) {
    // 残りの容量が足りず品物iを入れられないので、入れないパターンだけ処理
    // i+1 以降の品物のみを使ったときの最大値をそのままこの場合の最大値にする
    res = rec(i + 1, j);
  } else {
    // 品物iを入れるか入れないか選べるので、両方試して価値の和が大きい方を選ぶ
    res = max(
        rec(i + 1, j),
        rec(i + 1, j - w[i]) + v[i]
    );
  }
  return res;
}

void solve() {
  // 0番目以降で容量W以下の場合の結果を表示する
  cout << rec(0, W) << endl;
}

int main(void) {
  cin >> n >> W;
  for (long i=0; i<n; i++) {
    cin >> w[i] >> v[i];
  }
  solve();

  return 0;
}