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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define pb(a) push_back((a))
#define mp(a, b) make_pair((a), (b))
#define REP(a, b) for (ll i = (a); i < (b); i++)
#define REP2(a, b) for (ll j = (a); j < (b); j++)
#define REPE(a, b, c, d) \
    REP(a, b)            \
    for (ll j = (c); j < (d); j++)
#define REPV(a, b, c) for (ll(a) = b; (a) < (c); (a)++)
#define IREP(a, b) for (ll i = (a); i >= (b); i--)
#define IREP2(a, b) for (ll j = (a); j >= (b); j--)
#define IREPV(a, b, c) for (ll(a) = b; (a) >= (c); (a)--)
#define all(v) (v).begin(), (v).end()
#define trvs(c) for (auto &it : c)
#define INF 1e18
#define sz(x) x.size()
#define flash                         \
    ios_base::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0);
#define F first
#define S second
#define MOD 1000000007

using ll = long long;
using vll = vector<ll>;
using vvll = vector<vll>;
using pll = pair<ll, ll>;
using vpll = vector<pll>;
using vvpll = vector<vpll>;
#define _DEBUG 1
#if _DEBUG
#define debug(x) cout << #x << "=" << x << endl
#define debug2(x, y) cout << #x << "=" << x << " " << #y << "=" << y << endl;
#define debug3(x, y, z) cout << #x << "=" << x << " " << #y << "=" << y << " " << #z << "=" << z << endl;
#define debug4(x, y, z, w) cout << #x << "=" << x << " " << #y << "=" << y << " " << #z << "=" << z << " " << #w << "=" << w << endl;
#define debug5(x, y, z, w, t)      \
    cout << #x << "=" << x << " "; \
    debug4(y, z, w, t);
#define debug6(p, x, y, z, w, t)                            \
    cout << #p << " " << p << " " << #x << "=" << x << " "; \
    debug4(y, z, w, t);
#define debug7(o, p, x, y, z, w, t) \
    cout << #o << " " << o << " ";  \
    debug6(p, x, y, z, w, t);
#define wait  \
    int xxxx; \
    cin >> xxxx;
#define derr(x) cerr << #x << "=" << x << endl;
#define derr2(x, y) cerr << #x << "=" << x << " " << #y << "=" << y << endl;
#define derr3(x, y, z) cerr << #x << "=" << x << " " << #y << "=" << y << " " << #z << "=" << z << endl;
#define derr4(x, y, z, w) cerr << #x << "=" << x << " " << #y << "=" << y << " " << #z << "=" << z << " " << #w << "=" << w << endl;
#define derr5(x, y, z, w, t)       \
    cerr << #x << "=" << x << " "; \
    derr4(y, z, w, t);
#define derr6(p, x, y, z, w, t)                             \
    cerr << #p << " " << p << " " << #x << "=" << x << " "; \
    derr4(y, z, w, t);
#define derr7(o, p, x, y, z, w, t) \
    cerr << #o << " " << o << " "; \
    derr6(p, x, y, z, w, t);
ll checkpoint_counter = 0;
#define checkpoint cerr << "At checkpoint : " << checkpoint_counter++ << endl;

#else
#define debug(x) ;
#define debug2(x, y) ;
#define debug3(x, y, z) ;
#define debug4(x, y, z, q) ;
#define debug5(x, y, z, r, t) ;
#define debug6(x, y, z, r, t, s) ;
#define debug7(x, y, z, r, t, s, u) ;
#define wait ;
#define derr(x) ;
#define derr2(x, y) ;
#define derr3(x, y, z) ;
#define derr4(x, y, z, q) ;
#define derr5(x, y, z, r, t) ;
#define derr6(x, y, z, r, t, s) ;
#define derr7(x, y, z, r, t, f, u) ;
#define checkpoint ;
#endif

#define print_matrix(a, n, m) \
    REPE(0, n, 0, m) { cout << (a)[i][j] << ((j == m - 1) ? '\n' : ' '); }
#define present(container, element) (container.find(element) != container.end())
#define endl "\n"

auto cmp = [](pair<ll, pair<ll, ll>> a, pair<ll, pair<ll, ll>> b) {
    if (a.F == b.F)
        return a.S.F > b.S.F;
    return a.F > b.F;
};
//decltype(cmp)

void solve()
{
    ll n,w;cin>>n>>w;
    vll wei(n,0),cos(n,0);
    REP(0,n)cin>>wei[i]>>cos[i];
    vector<vpll> dp(n+1,vector<pll>(n,{0,0}));
    REPE(1,n+1,0,n){
       IREPV(l,i-1,0) REPV(k,0,n){
            if( k!=j and wei[j]+dp[l][k].S<=w)
            {
                if(dp[i][j].F< dp[l][k].F){
                    dp[i][j].F = dp[l][k].F;
                    dp[i][j].S = dp[l][k].S;
                }  
                else if(dp[i][j].F== dp[l][k].F){
                    dp[i][j].S = min(dp[i][j].S,dp[l][k].S+wei[j]);
                }   
            }
        }
        
        dp[i][j].F+=cos[j];
        dp[i][j].S+=wei[j];
        // if(j==n-1){
        //     for(auto xx : dp){
        //     for(auto yy : xx){
        //         cout<<"{"<<yy.F<<","<<yy.S<<"} ";
        //     }
        //     cout<<endl;
        // }cout<<endl;}
    }
    ll mx = -1;
    for(auto x : dp[n]){
        if(x.F>mx)mx = x.F;
    }
    cout<<mx<<endl;
}

int main()
{
    flash


    ll t;
    // cin >> t;
    // while (t--)
        solve();
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
    return 0;
}