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++14 (GCC 5.4.1) (judge verdict: Accepted)

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

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=(s);i<=(n);i++)
#define repr(i,n) for(int i=n-1;i>=0;i--)
#define REPR(i,s,n) for(int i=(s);i>=(n);i--)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define Eunique(v) v.erase(unique(all(v)),v.end())
#define Eback(s) s.erase(s.end()-1,s.end())
#define rev(v) reverse(all(v))
#define minvec(v) *min_element(all(v))
#define maxvec(v) *max_element(all(v))
#define sumvec(v) accumulate(all(v),0LL)
#define mapmin(v) v.rbegin()->first
#define mapmax(v) v.begin()->first
#define pb push_back
#define pf push_front
#define m_p make_pair
#define DOUBLE fixed << setprecision(15)
#define OK cerr<<"OK\n"
#define OK1 cerr<<"OK1\n"
#define OK2 cerr<<"OK2\n"
#define SIZE(s) (int)s.size()
#define INF ((1LL<<62)-(1LL<<31))
#define zero(x,n) setw(x) << setfill('0') << n
#define endl '\n'

typedef long long ll;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<long long> vll;
typedef vector<vll> vvll;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<char> vc;
typedef vector<vc> vvc;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<string> vs;

typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
typedef vector<pair<int,int>> vpii;
typedef vector<pair<ll,ll>> vpll;

const double pi = acos(-1.0);
const ll mod=1000000007;
const ll mod2=998244353;

template<class A, class B>
ostream& operator<<(ostream& ost, const pair<A, B>&p) {
	ost << "{" << p.first << ", " << p.second << "} ";
	return ost;
}

template<class T>
ostream& operator<<(ostream& ost, const vector<T>&v) {
	ost << "{";
	for (int i = 0; i<(int)v.size(); i++) {
		if (i)ost << " ";
		ost << v[i];
	}
	ost << "} \n";
	return ost;
}

template<class A, class B>
ostream& operator<<(ostream& ost, const map<A, B>&v) {
	ost << "{";
	for (auto p:v) {
		ost << "{" << p.first << ", " << p.second << "} ";
	}
	ost << "} ";
	return ost;
}

template<class T>
inline bool chmax(T& a, T b){if(a<b){a=b;return true;} return false;}

template<class T>
inline bool chmin(T& a, T b){if(a>b){a=b;return true;} return false;}

void YES(bool b) {cout << ((b)?"YES\n":"NO\n");}
void Yes(bool b) {cout << ((b)?"Yes\n":"No\n");}
void yes(bool b) {cout << ((b)?"yes\n":"no\n");}
void Yay(bool b) {cout << ((b)?"Yay!\n":":(\n");}

template<typename T>
void VIN(vector<T> &v,int n){for(int i=0;i<n;i++) cin >> v[i];}

template<typename T>
void VOUT(vector<T> &v){
    ll n=(int)v.size();
    for(int i=0;i<n;i++) cout << v[i] << (i==n-1?"\n":" ");
}

template<typename T>
void VOUTL(vector<T> &v){
    ll n=(int)v.size();
    for(int i=0;i<n;i++) cout << v[i] << endl;
}

ll powmod(ll a,ll b){ll c=1; while(b>0){ if(b&1){ c=a*c%mod; } a=a*a%mod; b>>=1; } return c;}
ll gcd(ll x, ll y) {return __gcd(x,y);}
ll lcm(ll x, ll y) {return x/__gcd(x,y)*y;}

int main()
{
	ll n,m;
	cin >> n >> m;
	vll w(n),v(n);
	rep(i,n) cin >> w[i] >> v[i];
	vvll dp(n+1,vll(100001,INF));
	dp[0][0]=0;
	for(ll i=1;i<=n;i++){
		rep(j,100001){
			if(j<v[i-1]) dp[i][j]=dp[i-1][j];
			else dp[i][j]=min(dp[i-1][j],dp[i-1][j-v[i-1]]+w[i-1]);
		}
	}
	for(ll i=100000;i>=0;i--){
		if(dp[n][i]<=m){
			cout << i << endl;
			return 0;
		}
	}
	return 0;
}