output_description: Print the maximum possible total value of the selected items.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N W
      w_1 v_1
      w_2 v_2
      :
      w_N v_N
attempt: import sys
  import heapq
  import re
  from itertools import permutations
  from bisect import bisect_left, bisect_right
  from collections import Counter, deque
  from fractions import gcd
  from math import factorial, sqrt, ceil
  from functools import lru_cache, reduce
  
  INF = 1 << 60
  MOD = 1000000007
  sys.setrecursionlimit(10**7)
  
  
  # UnionFind
  class UnionFind:
      def __init__(self, n):
          self.n = n
          self.parents = [-1] * n
  
      def find(self, x):
          if self.parents[x] < 0:
              return x
          else:
              self.parents[x] = self.find(self.parents[x])
              return self.parents[x]
  
      def union(self, x, y):
          x = self.find(x)
          y = self.find(y)
  
          if x == y:
              return
  
          if self.parents[x] > self.parents[y]:
              x, y = y, x
  
          self.parents[x] += self.parents[y]
          self.parents[y] = x
  
      def size(self, x):
          return -self.parents[self.find(x)]
  
      def same(self, x, y):
          return self.find(x) == self.find(y)
  
      def members(self, x):
          root = self.find(x)
          return [i for i in range(self.n) if self.find(i) == root]
  
      def roots(self):
          return [i for i, x in enumerate(self.parents) if x < 0]
  
      def group_count(self):
          return len(self.roots())
  
      def all_group_members(self):
          return {r: self.members(r) for r in self.roots()}
  
      def __str__(self):
          return "\n".join("{}: {}".format(r, self.members(r)) for r in self.roots())
  
  
  def is_prime(n):
      if n == 1:
          return False
      for i in range(2, int(n**0.5) + 1):
          if n % i == 0:
              return False
      return True
  
  
  # ワーシャルフロイド (任意の2頂点の対に対して最短経路を求める)
  # 計算量n^3 (nは頂点の数)
  def warshall_floyd(d, n):
      # d[i][j]: iからjへの最短距離
      for k in range(n):
          for i in range(n):
              for j in range(n):
                  d[i][j] = min(d[i][j], d[i][k] + d[k][j])
      return d
  
  
  # ダイクストラ
  def dijkstra_heap(s, edge, n):
      # 始点sから各頂点への最短距離
      d = [10**20] * n
      used = [True] * n  # True:未確定
      d[s] = 0
      used[s] = False
      edgelist = []
      for a, b in edge[s]:
          heapq.heappush(edgelist, a * (10**6) + b)
      while len(edgelist):
          minedge = heapq.heappop(edgelist)
          # まだ使われてない頂点の中から最小の距離のものを探す
          if not used[minedge % (10**6)]:
              continue
          v = minedge % (10**6)
          d[v] = minedge // (10**6)
          used[v] = False
          for e in edge[v]:
              if used[e[1]]:
                  heapq.heappush(edgelist, (e[0] + d[v]) * (10**6) + e[1])
      return d
  
  
  # 素因数分解
  def factorization(n):
      arr = []
      temp = n
      for i in range(2, int(-(-(n**0.5) // 1)) + 1):
          if temp % i == 0:
              cnt = 0
              while temp % i == 0:
                  cnt += 1
                  temp //= i
              arr.append([i, cnt])
  
      if temp != 1:
          arr.append([temp, 1])
  
      if arr == []:
          arr.append([n, 1])
  
      return arr
  
  
  # 2数の最小公倍数
  def lcm(x, y):
      return (x * y) // gcd(x, y)
  
  
  # リストの要素の最小公倍数
  def lcm_list(numbers):
      return reduce(lcm, numbers, 1)
  
  
  # リストの要素の最大公約数
  def gcd_list(numbers):
      return reduce(gcd, numbers)
  
  
  # 素数判定
  
  
  # limit以下の素数を列挙
  def eratosthenes(limit):
      A = [i for i in range(2, limit + 1)]
      P = []
  
      while True:
          prime = min(A)
  
          if prime > sqrt(limit):
              break
  
          P.append(prime)
  
          i = 0
          while i < len(A):
              if A[i] % prime == 0:
                  A.pop(i)
                  continue
              i += 1
  
      for a in A:
          P.append(a)
  
      return P
  
  
  # 同じものを含む順列
  def permutation_with_duplicates(L):
  
      if L == []:
          return [[]]
  
      else:
          ret = []
  
          # set（集合）型で重複を削除、ソート
          S = sorted(set(L))
  
          for i in S:
  
              data = L[:]
              data.remove(i)
  
              for j in permutation_with_duplicates(data):
                  ret.append([i] + j)
  
          return ret
  
  
  # ここから書き始める
  N, W = map(int, input().split())
  v = [[] for i in range(4)]
  w1 = 0
  for i in range(N):
      a, b = map(int, input().split())
      if i == 0:
          w1 = a
      v[a - w1].append(b)
  for i in range(4):
      v[i].sort(reverse=True)
  # print(v)
  ans = 0
  for i in range(N + 1):
      for j in range(N - i + 1):
          for k in range(N - i - j + 1):
              for l in range(N - i - j - k + 1):
                  if w1 * (i + j + k + l) + j + 2 * k + 3 * l > W:
                      continue
                  ans = max(
                      ans, sum(v[0][:i]) + sum(v[1][:j]) + sum(v[2][:k]) + sum(v[3][:l])
                  )
  print(ans)
  
problem_description: 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.
samples: [{"input": "4 6\n    2 1\n    3 4\n    4 10\n    3 4", "output": "11\n    \n\nThe first and third items should be selected.\n\n* * *"}, {"input": "4 6\n    2 1\n    3 7\n    4 10\n    3 6", "output": "13\n    \n\nThe second and fourth items should be selected.\n\n* * *"}, {"input": "4 10\n    1 100\n    1 100\n    1 100\n    1 100", "output": "400\n    \n\nYou can take everything.\n\n* * *"}, {"input": "4 1\n    10 100\n    10 100\n    10 100\n    10 100", "output": "0\n    \n\nYou can take nothing."}]