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: Wrong Answer)

#include<iostream>
#include<cassert>
#include<cctype>
#include<cerrno>
#include<cfloat>
#include<vector>
#include<algorithm>
#include<string>
#include<map>
#include<math.h>
#include<queue>
#include<deque>
#include<stack>
#include<cstdio>
#include<utility>
#include<set>
#include<list>
#include<cmath>
#include<stdio.h>
#include<string.h>
#include<iomanip>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<bitset>
#include<functional>
#include<iterator>
#include<complex>
#include<fstream>
#include<random>
#include<ctype.h>
#include<stdio.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define REP(i, n) FOR(i, 0, n - 1)
#define NREP(i, n) FOR(i, 1, n)
using ll = long long;
using pii = pair<int, int>;
using PII = pair<ll, ll>;
using piii = pair<pii, pii>;
using Pi = pair<int, pii>;
using Graph = vector<vector<int>>;
const int dx[4] = { 0, -1, 1, 0 };
const int dy[4] = { -1, 0, 0, 1 };
bool check(int x, int y) {
	if (0 <= x && x<55 && 0 <= y && y<55)return true;
	else return false;
}
const ll INF = 1e+7;
int gcd(int x, int y) {
	if (x < y)swap(x, y);
	if (y == 0)return x;
	return gcd(y, x%y);
}
void mul(ll a, ll b) {
	a = a * b % INF;
}
using Graph = vector<vector<int>>;

long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
const double PI = 3.14159265358979323846;
const int MAX = 510000;
ll fac[MAX], finv[MAX], inv[MAX];
//テーブルをつくる前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;//mod pにおける1,2,,,nの逆元
	for (int i = 2; i < MAX; ++i) {
		fac[i] = fac[i - 1] * i%INF;
		inv[i] = INF - inv[INF%i] * (INF / i) % INF;
		finv[i] = finv[i - 1] * inv[i] % INF;
	}
}

ll COM(int n, int k) {
	if (n < k)return 0;
	if (n < 0 || k < 0)return 0;
	return fac[n] * (finv[k] * finv[n - k] % INF) % INF;
}
double Euclidean_distance(double x1, double y1, double x2, double y2) {
	return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}

int prime[1001000];//i番目の素数
bool is_prime[1001010];//is_prime[i]がtrueならiは素数
					   //n以下の素数の数を返す
int sieve(int n) {
	int p = 0;
	for (int i = 0; i <= n; ++i) {
		is_prime[i] = true;
	}
	is_prime[0] = false;
	is_prime[1] = false;
	for (int i = 2; i <= n; ++i) {
		if (is_prime[i]) {
			prime[p] = i;
			p++;
			for (int j = 2 * i; j <= n; j += i) {
				is_prime[j] = false;
			}
		}
	}
	return p;
}

//素因数分解
map<int, int>prime_factor(int n) {
	map<int, int>res;
	for (int i = 2; i*i <= n; ++i) {
		while (n%i == 0) {
			++res[i];
			n /= i;
		}
	}
	if (n != 1)res[n] = 1;
	return res;
}


ll powmod(ll a, ll k,ll mod) {
	ll ap = a, ans = 1;
	while (k) {
		if (k & 1) {
			ans *= ap;
			ans %= mod;
		}
		ap = ap * ap;
		ap %= mod;
		k >>= 1;
	}
	return ans;
}
ll invi(ll a,ll mod) {
	return powmod(a, mod - 2,mod);
}
//逆元Aを足したときのmodで割った余りは
//+=invi(A)
///////////////////////////////////////

//dp[v][s]:=頂点vにいて所持金sという状態になるのに要する最小の所要時間
int main(){
	int n,W;
	cin>>n>>W;
	ll w[n+1],v[n+1];
	for(int i=1;i<=n;i++){
		cin>>w[i]>>v[i];
	}
	ll ans=0;
	ll dp[n+1][n+1]={0},A[n+1][n+1];
	for(int i=1;i<=n;i++){
		for(int j=2;j<=i;j++){
			for(int k=j-1;k<i;k++){
				if(A[k][j-1]+w[i]<=W&&dp[i][j]<dp[k][j-1]+v[i]){
					dp[i][j]=dp[k][j-1]+v[i];
					A[i][j]=A[k][j-1]+w[i];
				}
			}
			ans=max(ans,dp[i][j]);
		}
		dp[i][1]=v[i];
		A[i][1]=w[i];
		ans=max(ans,dp[i][i]);
	}

	cout<<ans;
}