// Slower C++ solution:
#include <cstdio>

#include <algorithm>

#include <vector>

using namespace std;

 

long long n,dp[100005],w,v,k,mxw,ans,pt,lpt;

bool d[100005];

int main(){

	scanf("%lld%lld",&n,&mxw);

	for (int i=100000;i>=1;i--){

		dp[i]=1e9+1;

	}

	d[0]=true; dp[0]=0;

	for (int i=1;i<=n;i++){

		scanf("%lld%lld",&w,&v); 

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

			if (d[j] == true && j+v <= 100000 && dp[j]+w <= mxw){

				d[j+v]=true;

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

			}

		}

	}

	for (int i=0;i<=100000;i++){

		if	(d[i])

		ans=i;

	}

	printf("%lld\n",ans);

	return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 1.5491424410390722x):
#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("input.txt", "r", stdin),freopen("output.txt", "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};

//int dx[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[] = {0 ,1, 1, 1, 0, -1, -1, -1};

// fflush(stdout)

// cout << flush

// a + b = a ^ b + 2 * (a & b)

int n, W;

ll w[101];

int v[101];

ll dp[2][N];

ll f(int i, int want){

    if(want < 0)return 1e18;

    if(i == n)return want == 0 ? 0 : 1e18;

    ll &ans = dp[i][want];

    if(ans == -1){

        ans = 1e18;

        ans = min(f(i + 1, want), ans);

        ans = min(ans, f(i + 1, want - v[i]) + w[i]);



    }return ans;

}

int main(){

    //FL;

    FIN;

    cin >> n >> W;

    // optimizando memoria

    int sum = 0;

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

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

        sum += v[i];

    }

    for(int i = n; i >= 0; i--){

        for(int want = 0; want <= sum; want++){// hast limite sumaMAx

            ll &ans = dp[i & 1][want];

            if(i == n){

                ans = want == 0 ? 0 : 1e18;

                continue;

            }

            ans = 1e18;

            ans = min(ans, dp[(i + 1) & 1][want]);

            if(want - v[i] >= 0)

                ans = min(ans, dp[(i + 1) & 1][want - v[i]] + w[i]);

        }

    }

    int ans = 0;

    for(int i = 0; i < sum + 1; i++){

        ll cur = dp[0 & 1][i];

        if(cur <= W)

            ans = i;

    }

    cout << ans << '\n';

    EjecuteTime;

    return 0;

}