// Slower C++ solution:
#include<bits/stdc++.h>



using namespace std;



#define all(v)				((v).begin()), ((v).end())

#define allr(v)				((v).rbegin()), ((v).rend())

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

#define ll long long

#define pb					push_back

#define fIO ios_base::sync_with_stdio(0);cin.tie(0);



const ll oo = 1e8;

const double pi = 3.1415926535897;





const double EPS = (1e-7);

int dcmp(double x, double y) {	return fabs(x-y) <= EPS ? 0 : x < y ? -1 : 1;	}





typedef vector<int>                vi;

typedef vector<pair<int,int>>      vii;

typedef vector<ll>                 vll;

typedef vector<double>             vd;

typedef vector< vi >               vvi;

typedef vector< vd >               vvd;

typedef vector<string>             vs;

typedef pair<int, int>             pii;



/**

    * PLEASEEE READDD PROBLEEEM CARFULLY ****

    * Always check for OVERFLOW ll vs int

    * Always check for array bounds

    * Check for function return

    * READ OUTPUT SECTION AGAIN!!!

**/



struct edge{

    int from, to, w;



    edge(){}

    edge(int x, int y, int we): from(x), to(y), w(we){}



    bool operator < (const edge &rhs) const{

        return w > rhs.w;

    }



};



const int N = 100 + 5, V = 1e5 + 5 ;



int dp[V];

int w[N], v[N];

int n, ans = 0;

//

//int sol(int i, int vS, int wS) {

//    if(i == n) {

//        ans = max(ans, vS);

//        return 0;

//    }

//

//    int &ret = dp[i][vS];

//    if(ret != -1) return ret;

//

//    ret = sol(i + 1, vS, wS);

//

//    if(wS - w[i] >= 0) ret = max(ret, sol(i + 1, vS + v[i], wS - w[i]));

//

//    return ret;

//}



int main(){

    fIO



//    freopen("", "r", stdin);

//    freopen("", "w", stdout);



    fill(dp, dp + V, 1e9 + 5);



    int cap;

    cin >> n >> cap;



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



    dp[0] = 0;

    for(int i = 0; i < n; i++) {

        for(int j = V - 5; j >= 1; j--) {

            if(j >= v[i]) {

               dp[j] = min(dp[j], dp[j - v[i]] + w[i]);

            }

        }

    }





    for(int i = V - 5; i >= 1; i--) {

        if(dp[i] <= cap) {

            cout << i << "\n";

            break;

        }

    }

}
















// Faster C++ solution for the same problem (measured speedup: 2.204226300253388x):
#include <bits/stdc++.h>



#define fst first

#define snd second

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

#define pb push_back

#define ALL(s) s.begin(),s.end()

#define FIN ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

#define SZ(s) int(s.size())

#define DBG(x) cerr <<#x << " = " << (x) << "\n"



#define SQR(x) ((x) * (x))



#define PI 3.1415926535897932385

#define INF 1000111222

#define eps 1e-7

#define maxN 11



using namespace std;



typedef long long ll;

typedef vector<ll> vll;

typedef pair<ll,ll> pll;

typedef pair<int,int> pii;

typedef vector<string> vstr;

typedef vector<vll> vvll;

typedef vector<bool> vbool;

typedef vector<pll> vpll;

typedef vector<vpll> vvpll;

typedef vector<vbool> vvbool;



int main(){FIN



	ll n,w,s=0;

	cin >> n >> w;

	vll ws(n),vs(n);

	fore(i,0,n){

		cin >> ws[i] >> vs[i];

		s+=vs[i];

	}

	vll dp(s+1,1e18L);

	dp[0] = 0;	

	fore(i,0,n){

		for(int j=(s-vs[i]);j>=0;j--)

			dp[j+vs[i]] = min(dp[j+vs[i]],dp[j]+ws[i]);

	}

	ll ans = 0;

	fore(i,0,s+1){

		if(dp[i]<=w) ans = max(ans,(ll)i);

	}

	cout << ans << "\n";

	return 0;

}