#include <bits/stdc++.h>
using ll = long long;
using dl = double long;
using namespace std;
const ll MOD = 1e9 + 7;

//From index 0 to i, what is the max value we can get with weight = w remaining?
ll f(int i, int w, vector<ll>& wt, vector<ll>& val, vector<vector<ll>>& dp){
    if(i == 0){
        if(wt[0] <= w) return val[0];
        else return 0;
    }
    if(dp[i][w] != -1) return dp[i][w];

    ll notTake = 0 + f(i - 1, w, wt, val, dp);
    ll take = INT_MIN;
    if(wt[i] <= w) take = val[i] + f(i - 1, w - wt[i], wt, val, dp);

    return dp[i][w] = max(take, notTake);
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    
    int n, W;
    cin >> n >> W;
    vector<ll> wt(n);
    vector<ll> val(n);
    for(int i = 0; i < n; i++){
        cin >> wt[i] >> val[i];
    }

    vector<vector<ll>> dp(n + 1, vector<ll> (W + 1, -1));
    ll ans = f(n - 1, W, wt, val, dp);

    cout << ans << endl;

    return 0;
}

#include <bits/stdc++.h>
using ll = long long;
using dl = double long;
using namespace std;
const ll MOD = 1e9 + 7;

//From index 0 to i, to achieve value = v what is the min weight required?
ll f(ll i, ll v, vector<ll>& wt, vector<ll>& val, vector<vector<ll>>& dp){
    if(i == 0){
        if(v == 0) return 0;
        if(val[0] == v) return wt[0];
        else return 1e15;
    }
    if(dp[i][v] != -1) return dp[i][v];

    ll notTake = 0 + f(i - 1, v, wt, val, dp);
    ll take = INT_MAX;
    if(val[i] <= v) take = wt[i] + f(i - 1, v - val[i], wt, val, dp);

    return dp[i][v] = min(take, notTake);
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    
    ll n, W;
    cin >> n >> W;
    vector<ll> wt(n);
    vector<ll> val(n);
    int V = 0;
    for(int i = 0; i < n; i++){
        cin >> wt[i] >> val[i];

        V += val[i];
    }

    vector<vector<ll>> dp(n + 1, vector<ll> (V + 1, -1));

    for(ll v = V; v >= 0; v--){
        if(f(n - 1, v, wt, val, dp) <= W){
            cout << v << endl; //First value from maximum which can be possible within given max. wt(W) of knapsack
            break;
        }
    }

    return 0;
}