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>
#define FIN         ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ii          pair<int,int>
#define F           first
#define S           second
#define pb          push_back
#define all(x)      (x).begin(),(x).end()
#define rall(x)     (x).rbegin(),(x).rend()
#define fore(i,a,b) for(int i = a;i < b; i+= 1)
#define forr(i,a)   for(int i = a; i >= 0; i--)
#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 
#define w(t)        while(t--)
#define sz(s)       int(s.size())
#define cls(a,car)  memset(a,car,sizeof (a))
#define db(x)       cerr << #x << " is " << x << '\n'
#define FL          freopen("in", "r", stdin),freopen("out", "w", stdout);
#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"
#define angle(x)    double(x * acos(-1) / 180.0)
using namespace std;
typedef long long   ll;
typedef vector<int> vi;
typedef vector<ii>  vii;
const int N = 1 * 1e5 + 5;
const ll mod = 1e9 + 7;
const double E = 1e-9;
int dx[] = {1, 0}, dy[] = {0, 1};
//int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};
// fflush(stdout)
// cout << flush
// a + b = a ^ b + 2 * (a & b)
int n;
ll W;
ll dp[101][100010];
ll w[101], v[101];
ll f(int i, int want){
    if(i == n){
        return want == 0 ? 0: 1e16;
    }
    ll &ans = dp[i][want];
    if(ans != -1)return ans;
    ans = 1e17;
    ans = min(f(i + 1, want - v[i]) + w[i], ans);
    ans = min(f(i + 1, want), ans);
    return ans;
} 
int main(){
    //FL;
    FIN;
    cin >> n >> W;
    ll sum = 0;
    fore(i, 0, n){
        cin >> w[i] >> v[i];
        sum += v[i];
    }
    cls(dp, -1);
    int ans = 0;
    fore(i, 0, 100010){
        ll cur = f(0, i);
        if(cur <= W)
            ans = i;
    }
    cout << ans << '\n';
    EjecuteTime;
    return 0;
}