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: Runtime Error)

// code written by kshitij_coder
#include<bits/stdc++.h>
using namespace std;
 
#define ll int64_t
#define pii pair<ll,ll>
#define psi pair<string,ll> 
#define pb push_back
#define mp make_pair
#define MOD 1000000007
#define pb push_back
#define lim 100000
#define f(k) k.first
#define s(k) k.second
#define bcnt __builtin_popcount
#define endl "\n"
#define PI 3.1415926535

struct hash_pair { 
    template <class T1, class T2> 
    size_t operator()(const pair<T1, T2>& p) const{ 
        auto hash1 = hash<T1>{}(p.first); 
        auto hash2 = hash<T2>{}(p.second); 
        return hash1 ^ hash2; 
    } 
}; 

ll power(ll a,ll b){ ll ans = 1; ll k = a;while(b > 0){ if(b & 1){ ans *= k; ans %= MOD; } k = (k*k)%MOD; b = b >> 1;} return ans;}

// ll ncr(ll n,ll r,ll*arr){
//     if(r==0) return 1;
//     ll res = 1;
//     return (arr[n]*(power(arr[n-r],MOD-2))%MOD*(power(arr[r],MOD-2))%MOD)%MOD;
// }

ll gcd(ll a,ll b){ if(a == 0) return b; if(b == 0) return a; return gcd(b,a%b); }
ll add_mod(ll a,ll b,ll m){return ((a%m) + (b%m))%m;}
ll sub_mod(ll a,ll b,ll m){return ((a%m) - (b%m) + m)%m;}
ll mul_mod(ll a,ll b,ll m){return ((a%m)*(b%m))%m;}
void print(double val,int k){ cout<<fixed<<setprecision(k)<<val<<endl;}
ll toint(string s) {stringstream ss; ss << s; ll x; ss >> x; return x;}

void solve(){
    int n,w;
    cin >> n >> w;
    
    ll **dp = new ll*[n + 1];
    for(int i = 0;i <= n;i++) dp[i] = new ll[w+1];
    
    for(int i = 0;i <= n;i++) dp[i][0] = 0;
    for(int i = 0;i <= w;i++) dp[0][i] = 0;
    
    for(int i = 1;i<=n;i++){
        int u,v;
        cin >> u >> v;
        
        for(int j = 1; j <= w;j++){
            ll a = ((j - u) >= 0) ? (dp[i-1][j-u] + v) : 0;
            ll b = dp[i-1][j];
            dp[i][j] = max(a,b);
        }
    }
    
    
    cout<<dp[n][w]<<endl;
}

int main(){
    ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	solve();
    return 0;
}