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

#include<cstdio>

#include<algorithm>



using namespace std;



const int INF = 1e9 + 7;

const int MAX_V = 1e3 + 1;

const int MAX_N = 101;

int max_V;

int n,W;

int w[MAX_N],v[MAX_N];

int dp[MAX_N+1][MAX_N * MAX_V+1];



//dp[i+1][j]:从前i件物品中得到价值为j时的最小重量,不存在就是INF

//转移的话,就看第i类是要还是不要

int solve(){

    fill(dp[0], dp[0] + MAX_V * MAX_N + 1, INF);//没有物品可拿,自然是INF

    dp[0][0] = 0;

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

        for (int j = 0; j <= max_V * n; j++){

            if(j<v[i]){

                dp[i + 1][j] = dp[i][j];//装不下

            }else{

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

            }

        }

    }

    int res = 0;

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

        

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

            res = i;

            //cout << dp[n][i] << " ";

        }

        

    }

    return res;

}



int main(){

    scanf("%d%d", &n, &W);

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

        scanf("%d %d", &w[i], &v[i]);

        max_V = max(max_V, v[i]);

    }

    cout << solve() << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.356703615880905x):
#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[101][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;

    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][want];

            if(i == n){

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

                continue;

            }

            ans = 1e18;

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

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

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

        }

    }

    int ans = 0;

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

        ll cur = dp[0][i];

        if(cur <= W)

            ans = i;

    }

    cout << ans << '\n';

    EjecuteTime;

    return 0;

}