output_description: Print the maximum possible sum of the values of items that Taro takes home.
  
  * * *
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 heapq import heapify, heappop, heappush
  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())
  
  
  # 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
  def is_prime(n):
      if n <= 1:
          return False
      p = 2
      while True:
          if p**2 > n:
              break
          if n % p == 0:
              return False
          p += 1
      return True
  
  
  # 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())
  dp = [[INF for j in range(10**5 + 1)] for i in range(n)]
  for i in range(n):
      weight, value = map(int, input().split())
      if i == 0:
          dp[i][value] = weight
          continue
      for j in range(10**5 + 1):
          dp[i][j] = dp[i - 1][j]
          if 0 <= j - value:
              dp[i][j] = min(dp[i][j], dp[i - 1][j - value] + weight)
  ans = 0
  # print(dp)
  for i in reversed(range(10**5 + 1)):
      if dp[-1][i] <= w:
          ans = i
          break
  print(ans)
  
problem_description: Statement
  
  There are N items, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N),
  Item i has a weight of w_i and a value of v_i.
  
  Taro has decided to choose some of the N items and carry them home in a
  knapsack. The capacity of the knapsack is W, which means that the sum of the
  weights of items taken must be at most W.
  
  Find the maximum possible sum of the values of items that Taro takes home.
samples: [{"input": "3 8\n    3 30\n    4 50\n    5 60", "output": "90\n    \n\nItems 1 and 3 should be taken. Then, the sum of the weights is 3 + 5 = 8, and\nthe sum of the values is 30 + 60 = 90.\n\n* * *"}, {"input": "1 1000000000\n    1000000000 10", "output": "10\n    \n\n* * *"}, {"input": "6 15\n    6 5\n    5 6\n    6 4\n    6 6\n    3 5\n    7 2", "output": "17\n    \n\nItems 2, 4 and 5 should be taken. Then, the sum of the weights is 5 + 6 + 3 =\n14, and the sum of the values is 6 + 6 + 5 = 17."}]