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 <bits/stdc++.h>
 
using namespace std;
 
#define PB push_back
#define MP make_pair
#define INS insert
#define LB lower_bound
#define UB upper_bound
#define pii pair <int,int>
#define pll pair <long long, long long>
#define si pair<string, int>
#define is pair<int, string>
#define X first
#define Y second
#define _ << " " <<
#define sz(x) (int)x.size()
#define all(a) (a).begin(),(a).end()
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORD(i, a, b) for (int i = (a); i > (b); --i)
#define FORR(i, l, r) for (int i = (l); i <= (r); ++i)
#define FORP(i, a, b) for ((i) = (a); (i) < (b); ++i)
#define FORA(i, x) for (auto &i : x)
#define REP(i, n) FOR(i, 0, n)
#define BITS(x) __builtin_popcount(x)
#define MSET memset
#define MCPY memcpy
#define SQ(a) (a) * (a)
 
typedef long long ll;
typedef long double ld;
typedef vector <int> vi;
typedef vector <pii> vpi;
typedef vector <ll> vll;
typedef vector <pll> vpl;
typedef vector <double> vd;
typedef vector <ld> vld;
typedef vector<si> vsi;
typedef vector<is> vis;
typedef vector<string> vs;
//((float) t)/CLOCKS_PER_SEC
 
const int MOD = 1e9 + 7;
const int PI = acos(-1);
const int LOG = 20;
const int INF = 1e9 + 10;
const ll INFL = 1e18 + 10;
const int ABC = 30;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
const int dox[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int doy[] = {0, 0, -1, 1, -1, 1, -1, 1};
 
inline int sum(int a, int b){
  if (a + b >= MOD)
  	return a + b - MOD;
  if (a + b < 0)
  	return a + b + MOD;
  return a + b;
}
 
inline void add(int &a, int b){
  a = sum(a, b);
}
 
inline int pot(ll pot, int n){
	ll ret = 1;
	while (n){
		if (n & 1)
			ret = (ret * pot) % MOD;
		pot = (pot * pot) % MOD;
		n >>= 1;
	}
	return ret;
}
 
inline int mul(int a, int b){
  return (ll)a * (ll)b % MOD;
}
 
inline int sub(int a, int b){
	return (a - b + MOD) % MOD;
}
 
inline int divide(int a, int b){
	return mul(a, pot(b, MOD - 2));
}
 
ll lcm(ll a, ll b){
	return abs(a * b) / __gcd(a, b);
}
 
inline double ccw(pii A, pii B, pii C){
	return (A.X * B.Y) - (A.Y * B.X) + (B.X * C.Y) - (B.Y * C.X) + (C.X * A.Y) - (C.Y * A.X);
}
 
inline int CCW(pii A, pii B, pii C){
	double val = ccw(A, B, C);
	double eps = max(max(abs(A.X), abs(A.Y)), max(max(abs(B.X), abs(B.Y)), max(abs(C.X), abs(C.Y)))) / 1e9;
	if (val <= -eps)
		return -1;
	if (val >= eps)
		return 1;
	return 0;
}
 
void to_upper(string &x){
	REP(i, sz(x))
		x[i] = toupper(x[i]);
}
 
void to_lower(string &x){
	REP(i, sz(x))
		x[i] = tolower(x[i]);
}
 
string its(ll x){
	if (x == 0)
		return "0";
	string ret = "";
	while (x > 0){
		ret += (x % 10) + '0';
		x /= 10;
	}
	reverse(all(ret));
	return ret;
}
 
ll sti(string s){
	ll ret = 0;
	REP(i, sz(s)){
		ret *= 10;
		ret += (s[i] - '0');
	}
	return ret;	
}
 
const int N = 100;
const int W = 1e5 + 10;
const int OFF = (1 << LOG); 

ll memo[N][W], n, w;
vector <vll> v;

ll dp(int x, int c, ll wi){
	if (x == n)
		return c;
	auto &ret = memo[x][c];
	if (~ret)
		return ret;
	if (wi + v[x][0] <= w)
		ret = dp(x + 1, c + v[x][1], wi + v[x][0]);
	return ret = max(ret, dp(x + 1, c, wi));
}

int main ()	{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	cin >> n >> w;
	REP(i, n){
		ll x, y; cin >> x >> y;
		v.PB({x, y});
	}
	MSET(memo, -1, sizeof memo);
	cout << dp(0, 0, 0) << '\n';
 
	return 0;
}