Problem:
Score : 100 points

Problem 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.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq W \leq 10^9

1 \leq w_i \leq W

1 \leq v_i \leq 10^3

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 sum of the values of items that Taro takes home.

Sample Input 1

3 8
3 30
4 50
5 60

Sample Output 1

90

Items 1 and 3 should be taken.
Then, the sum of the weights is 3 + 5 = 8, and the sum of the values is 30 + 60 = 90.

Sample Input 2

1 1000000000
1000000000 10

Sample Output 2

10

Sample Input 3

6 15
6 5
5 6
6 4
6 6
3 5
7 2

Sample Output 3

17

Items 2, 4 and 5 should be taken.
Then, the sum of the weights is 5 + 6 + 3 = 14, and the sum of the values is 6 + 6 + 5 = 17.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
	*#*#*#* Author GaryMr *#*#*#*
########################################################################
	是否可以简化问题：
	dfs or bfs
	是否可以化具体为抽象
	字符串问题是否可以用数学模拟递归过程（取c[i]或不取来算结果）
	二分？左闭右开？
	long long？
	逆向思维？
	dp？
	贪心？
	匹配括号 左减右>=0
	多组数据清空？
	dp优化用g？
	线段树：大小为n的8倍
	线段树：左闭右开！
	线段树：find： if(l>=b||r<=a) return 0x3f3f3f3f; if(r<=b&&l>=a) return tree[k];
	线段树：满二叉树 
	循环跳出条件 
	代数设出每个值寻找关系？
	优先队列：priority_queue<类型> q;   
	priority_queue<类,vector<类>, less<类> > pq1; 　　　 // 使用递减	
	priority_queue<类，vector<类>, greater<类> > pq2; 　　// 使用递增
	判断括号，判断是否长度为偶数
	1<<ceil(log2(double(n)))
*/
#include<bits/stdc++.h>
#define rb(a,b,c) for(int a=b;a<=c;a++)
#define rl(a,b,c) for(int a=b;a>=c;a--)
#define niv vector<int>
#define LL long long
#define IT iterator
#define FIR first
#define SEC second
using namespace std;
const int MAX=0x3f3f3f3f;
typedef pair<int,int> mp;
typedef pair<mp,mp> superpair;
const int MAXVAL=100001;
LL dp[MAXVAL+5];
int main(){
	rb(i,1,MAXVAL) dp[i]=9000000000;dp[0]=0;
	LL n,w;
	scanf("%I64d %I64d",&n,&w);
	rb(i,1,n){
		int a,b;
		scanf("%d %d",&a,&b);
		rl(j,MAXVAL,b){
			dp[j]=min(dp[j],dp[j-b]+a);
		}
	}
	int final;
	rb(i,0,MAXVAL){
		if(dp[i]<=w){
			final=i;
		}
	}
	printf("%d\n",final);
	return 0;
}