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: Time Limit Exceeded)

//author:satwik_bhv1
#include<bits/stdc++.h>
//datatypes
#define ll long long 
#define ld long double
//loops
#define fr(i,z,n) for(ll i=z;i<n;i++)
#define br(i,z,n) for(ll i=z;i>n;i--)
//operations
#define mp  make_pair
#define ff  first
#define ss  second
#define pub push_back
#define all(v) v.begin(),v.end()
//map
#define mi map<ll,ll>
//vectors
#define vi vector<ll>
#define vpi vector<pair<ll,ll> >
//constants
#define pi 3.1415926535897932384626
#define mod 1000000007 
#define MAXN 100001
/*notes
    Don't use inbuilt fun for power
    Range of longlong=(-9,223,372,036,854,775,808 to 9,223,372,036,854,775,807 )
    INT_MAX = 2147483647(aprox 2*10^10)
    INT_MIN = -2147483648
    LLONG_MAX = 9223372036854775807 
    LLONG_MIN = -9223372036854775808
    __gcd is the function in built for hcf
    priority queue is max heap by default
    topological sort -> khan's algo use bfs and indegree of node  
    sssp(positive weighted edges) -> Dijkstras
    apsp -> Floydwarshall
    MST -> kruskal(DSU)
*/
using namespace std;
//functions
ll power(ll x,ll y){ll res=1;while(y){if(y%2){res*=x;}y/=2;x*=x;} return res;}
ll power(ll x,ll y,ll z){ll res=1;while(y){if(y%2){res=(res*x)%z;}y/=2;x=(x*x)%z;} return res;}
void fast() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); }
bool sortbysec(const pair<ll, ll>& a, const pair<ll, ll>& b) { return (a.second > b.second); }
ll ceil(ll x,ll y){return (x/y)+(x%y!=0);}
ll n,w;
ll v[101][2];
map<ll,ll> dp[101];
ll fun(ll pos,ll limit)
{
    if(pos==n)
    {
        return 0;
    }
    ll ans=0;
    if(dp[pos+1].find(limit)==dp[pos+1].end())
    {
        ans=max(ans,fun(pos+1,limit));
    }
    else
    {
        ans=max(ans,dp[pos+1][limit]);
    }
    if(limit>=v[pos][0])
    {
        if(dp[pos+1].find(limit-v[pos][0])==dp[pos+1].end())
        {
            ans=max(ans,fun(pos+1,limit-v[pos][0])+v[pos][1]);
        }
        else
        {
            ans=max(ans,dp[pos+1][limit-v[pos][0]]+v[pos][1]);
        }
    }
    return dp[pos][limit]=ans;
}
void solve()
{
    cin>>n>>w;
    fr(i,0,n)
    {
        fr(j,0,2)
        {
            cin>>v[i][j];
        }
    }
    cout<<fun(0,w);
}
int main()
{
    fast();
    ll t;
    t=1;
    //cin>>t;
    fr(i,0,t)
    {
        solve();
    }
}
/*





























*/