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: Accepted)

#include    <bits/stdc++.h>
#include    <ext/pb_ds/assoc_container.hpp>
#include    <ext/pb_ds/trie_policy.hpp>
#include    <ext/rope>

#define     ll              long long
#define     ld              long double
#define     fi              first
#define     se              second
#define     sz(x)           x.size()
#define     mp              make_pair
#define     pb              push_back
#define     eb              emplace_back
#define     all(x)          begin(x), end(x)
#define     rall(x)         rbegin(x), rend(x)
#define     rep(i, x, n)    for (ll i = x; i < n; i++)
#define     fastio()        ios_base::sync_with_stdio(NULL); cin.tie(0); cout.tie(0)
#define     int             long long
using       namespace std;
using       namespace __gnu_pbds;
using       namespace __gnu_cxx;

mt19937     rng(chrono::steady_clock::now().time_since_epoch().count());

typedef     vector<int>     vi;
typedef     vector<vi>      vvi;
typedef     pair<int, int>  ii;
typedef     vector<ii>      vii;
typedef     vector<ll>      vl;
typedef     vector<vl>      vvl;
typedef     pair<ll, ll>    pll;
typedef     vector<pll>     vll;

inline void EXECUTE_TIME() {cout << endl << setprecision(10) << fixed << "CPU Time: " << clock() / double(CLOCKS_PER_SEC) << " sec(s)\n";}

template <class T> inline T RD(T &a) {T x; cin >> x; return a = x;}
template <class T> inline T gcd(T a, T b) {if (b) return gcd(b, a % b); else return a;}
template <class T> inline T lcm(T a, T b) {return a / gcd(a,b) * b;}
template<typename T> using ordered_set=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef trie<string, null_type, trie_string_access_traits<>, pat_trie_tag, trie_prefix_search_node_update> pref_trie;

const int maxn = 105, maxw = 1e5 + 5, inf = 1e12;

int dp[maxn][maxw];
int w[maxn], v[maxn];

signed main()
{
//    freopen("test.inp","r",stdin);
//    freopen("test.out","w",stdout);
	fastio();

    ll n, W;
	cin >> n >> W;
	for (ll i = 0; i < n; i++) cin >> w[i] >> v[i];

	for (ll i = 0; i < maxn; i++){
		for (ll j = 0; j < maxw; j++) {
            dp[i][j] = inf;
		}
	}

	for (ll i = 0; i < maxw; i++) dp[0][i] = inf;

	dp[0][0] = 0;

	for (ll i = 1; i <= n; i++){
		for (ll j = 0; j < maxw; j++){
			if (j + v[i-1] < maxw) dp[i][j + v[i - 1]] = min(dp[i][j + v[i - 1]], dp[i - 1][j] + w[i - 1]);
			dp[i][j] = min(dp[i][j], dp[i-1][j]);
		}
	}

	ll mx = 0;

	for (ll j = 0; j < maxw; j++){
		for (int i = 0; i <= n; i++){
			if (dp[i][j] <= W) mx = max(mx, j);
		}
	}

	cout << mx;
}