// Original C++ solution (slower):
// includes

#include <cstdio>

#include <cstdint>

#include <iostream>

#include <iomanip>

#include <string>

#include <queue>

#include <stack>

#include <vector>

#include <set>

#include <map>

#include <unordered_map>

#include <algorithm>

#include <utility>

#include <functional>

#include <cmath>

#include <climits>

#include <bitset>



// macros

#define ll long long int

#define pb push_back

#define mk make_pair

#define pq priority_queue

#define FOR(i, a, b) for(int i=(a); i<(b);++i)

#define rep(i, n) FOR(i, 0, n)



using namespace std;



//  types

typedef pair<int, int> P;

typedef pair<ll, int> Pl;

typedef pair<ll, ll> Pll;

typedef pair<double, double> Pd;

 

// constants

const int inf = 1e9;

const ll linf = 1LL << 50;

const double EPS = 1e-10;



// solve

ll dp[101][100001];



int main(int argc, char const* argv[])

{

  int n;

  ll W;

  cin >> n >> W;

  vector<ll> w(n);

  vector<int> v(n);

  rep(i, n){

    cin >> w[i] >> v[i];

  }

  rep(i, n+1){

    rep(j, 100001){

      dp[i][j] = linf / 2;

    }

  }

  dp[0][0] = 0;

  rep(i, n){

    rep(j, 100001){

      dp[i+1][j] = dp[i][j];

      if(j-v[i]>=0)dp[i+1][j] = min(dp[i+1][j], dp[i][j-v[i]]+w[i]);

    }

  }

  for(int j = 100000; j >= 0; j--){

    if(dp[n][j] <= W){

      cout << j << endl;

      return 0;

    }

  }

	return 0;

}


// Optimized solution to the same problem (faster):
// includes

#include <bits/stdc++.h>

using namespace std;



// macros

#define pb emplace_back

#define mk make_pair

#define FOR(i, a, b) for(int i=(a);i<(b);++i)

#define rep(i, n) FOR(i, 0, n)

#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)

#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)

#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)

#define whole(x) (x).begin(),(x).end()

#define sz(x) ((int)(x).size())

#define bit(n) (1LL<<(n))

#define F first

#define S second

// functions

template <typename T> void unique(T& c){c.erase(std::unique(c.begin(), c.end()), c.end());}

template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}

template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}

template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}

template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}

template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}

template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}

template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}

template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}

template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << "(" << p.first << ", " << p.second << ")"; return os;}

template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << "(" << itr->first << ", " << itr->second << ")"; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}

template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << "(" << itr->first << ", " << itr->second << ")"; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}

//  types

using ll = long long int;

using P = pair<int, int>;

// constants

const int inf = 1e9;

const ll linf = 1LL << 50;

const double EPS = 1e-10;

const int mod = 1000000007;

const int dx[4] = {-1, 0, 1, 0};

const int dy[4] = {0, -1, 0, 1};

// io

struct fast_io{

  fast_io(){ios_base::sync_with_stdio(false); cin.tie(nullptr); cout << fixed << setprecision(20);}

} fast_io_;



const int B = 100000;

ll dp[B+10];



int main(int argc, char const* argv[])

{

  int n; cin >> n;

  ll W; cin >> W;

  vector<ll> w(n), v(n); rep(i, n)cin >> w[i] >> v[i];

  rep(i, B + 1)dp[i] = linf;



  dp[0] = 0;

  rep(i, n){

    for(int j = B; j >= 0; j--){

      if(j-v[i]>=0)chmin(dp[j], dp[j-v[i]] + w[i]);

    }

  }



  ll res = 0;

  rep(i, B + 1){

    if(dp[i] <= W)res = i;

  }

  cout << res << endl;

  return 0;

}


// Fastest known solution for this problem: