stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
//by Top1

#include <bits/stdc++.h>

     

#define f first

#define s second

#define int long long

#define ld long double

#define ull unsigned long long

#define pii pair <int, int>

#define pll pair <long, long>

#define mp make_pair

#define pb push_back

#define pf push_front

#define pp pop_back

#define rev reverse

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

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

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

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

#define nxtp next_permutation

#define pvtp prev_permutation

#define KZ return 

#define x1 x21212121

#define y1 y21212121

#define sqr(x) x * x 

 

const int inf = (1e9) + 1;

const int mod = 1e9 + 7;

const int maxn = 1e5 + 111;

     

using namespace std;



int q, n, w, a[maxn], b[maxn], dp[105][maxn], ans;



void solve(){    

	cin >> n >> w;



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

		cin >> a[i] >> b[i];



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

		for (int j = 1e5; j >= 0; j--)

			dp[i][j] = inf;



	dp[0][0] = 0;



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

		for (int j = 1e5; j >= 0; j--){

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

			if (j - b[i] >= 0)

				dp[i][j] = min(dp[i][j], dp[i - 1][j - b[i]] + a[i]);

			if (dp[i][j] <= w)

				ans = max(ans, j);

		}

		/*for (int j = 1; j <= 20; j++)

			cout << dp[i][j] << " ";

		cout << endl;	*/

		//cout << dp[i][100000] << endl;

	}



	//cout << dp[n][100000] << " ";



	/*for (int i = 0; i <= w; i++)

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

	*/

	cout << ans;

}



main(){

    boost;

	    

	q = 1;



	while (q--){

		solve();

	}

}

// Faster C++ solution for the same problem (measured speedup: 2.834568720827436x):
// @author: Amirul Islam



#include <bits/stdc++.h>

using namespace std;

#define LLI long long int

#define PII pair <int, int>



const int mx = 1e5+5;

const LLI INF = 1e10;



LLI dp[mx], w, val, N, W;



LLI max(LLI a, LLI b) {

    return (a > b ? a : b);

}



LLI knapsack(int N, int W) {

    for (int i = 0; i < mx; i++) dp[i] = INF;

    dp[0] = 0;

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

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

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

            if (j >= val) dp[j] = min(dp[j], dp[j-val] + w);

        }

    }

    LLI ret = 0;

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

        if (dp[i] <= W) ret = max(ret, i);

    }

    return ret;

} 



int main() {

    scanf("%lld %lld", &N, &W);

    printf("%lld\n", knapsack(N, W));

    return 0;

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

using namespace std;

typedef long long int ll;

const int MAX=1e5+10;

ll dp[MAX];

int main (){

    int N,W;cin >> N >> W;

    memset(dp,63,sizeof dp);

    dp[0]=0;

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

        int w,val;

        cin >> w >> val;

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

            if(j-val>=0)

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

        }



    }

    int res=0;

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

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

    }

    cout<<res<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.1599358265364155x):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

const int N=1005;

const int maxn=1e5+5;

ll a[N],v[N];

ll dp[maxn];

int main(){

	int n,k;

	cin>>n>>k;

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

		cin>>a[i]>>v[i];

	memset(dp,0x3f,sizeof(dp));

	dp[0]=0;

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

		for(int j=maxn-5;j>=v[i];--j)

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

	ll ans=0;

	for(int i=1;i<=maxn-5;++i)

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

	cout<<ans<<endl;

	return 0;

}
// Slower C++ solution:
//by Top1

#include <bits/stdc++.h>

     

#define f first

#define s second

#define int long long

#define ld long double

#define ull unsigned long long

#define pii pair <int, int>

#define pll pair <long, long>

#define mp make_pair

#define pb push_back

#define pf push_front

#define pp pop_back

#define rev reverse

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

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

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

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

#define nxtp next_permutation

#define pvtp prev_permutation

#define KZ return 

#define x1 x21212121

#define y1 y21212121

#define sqr(x) x * x 

 

const int inf = (1e9) + 1;

const int mod = 1e9 + 7;

const int maxn = 1e5 + 111;

     

using namespace std;



int q, n, w, a[maxn], b[maxn], dp[105][maxn], ans;



void solve(){    

	cin >> n >> w;



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

		cin >> a[i] >> b[i];



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

		for (int j = 1e5; j >= 0; j--)

			dp[i][j] = inf;



	dp[0][0] = 0;



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

		for (int j = 1e5; j >= 0; j--){

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

			if (j - b[i] >= 0)

				dp[i][j] = min(dp[i][j], dp[i - 1][j - b[i]] + a[i]);

			if (dp[i][j] <= w)

				ans = max(ans, j);

		}

		/*for (int j = 1; j <= 20; j++)

			cout << dp[i][j] << " ";

		cout << endl;	*/

		//cout << dp[i][100000] << endl;

	}



	//cout << dp[n][100000] << " ";



	/*for (int i = 0; i <= w; i++)

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

	*/

	cout << ans;

}



main(){

    boost;

	    

	q = 1;



	while (q--){

		solve();

	}

}

// Faster C++ solution for the same problem (measured speedup: 2.8153596448417493x):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

const int N=1005;

const int maxn=1e5+5;

ll a[N],v[N];

ll dp[maxn];

int main(){

	int n,k;

	cin>>n>>k;

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

		cin>>a[i]>>v[i];

	memset(dp,0x3f,sizeof(dp));

	dp[0]=0;

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

		for(int j=maxn-5;j>=v[i];--j)

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

	ll ans=0;

	for(int i=1;i<=maxn-5;++i)

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

	cout<<ans<<endl;

	return 0;

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

using namespace std;

typedef long long int ll;

const int MAX=1e5+10;

ll dp[MAX];

int main (){

    int N,W;cin >> N >> W;

    memset(dp,63,sizeof dp);

    dp[0]=0;

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

        int w,val;

        cin >> w >> val;

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

            if(j-val>=0)

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

        }



    }

    int res=0;

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

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

    }

    cout<<res<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.6766726508151812x):
#include <iostream>

#include <cstring>

using namespace std;

const int maxn=1e5+5;

using ll = long long;

ll dp[maxn];

ll w[maxn],v[maxn];

int main()

{

   ll n,m,mm;

   cin>>n>>mm;

   m=0;

   memset(dp,0x3f,sizeof(dp));

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

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

       m+=v[i];

   }

   dp[0]=0;

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

       for(ll j=m;j>=0;--j){

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

          // cout<<dp[j]<<endl;

       }

   }

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

       if(dp[i]<=mm){

           cout<<i<<endl;

           break;

       }

   }

   return 0;

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

using namespace std;



#define pb push_back

#define mk make_pair

#define fi first

#define se second

#define eb emplace_back



typedef long long ll;

typedef pair<int,int> ii;

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

const ll INF = 0x3f3f3f3f3f3fef;



const int V = 1e5+2;

const int N = 102;



ll dp[V];

ll p[N];

ll v[N];

ll n,w;

 

int main() {

    ios_base::sync_with_stdio(false);

    for(int i = 0; i < V; i++) dp[i] = INF;



    cin >> n >> w;

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

    

    dp[0] = 0;



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

        for(ll j = V-1; j-v[i] >= 0; j--)

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



    for(int i = V-1; i >= 0; i--)

        if(dp[i] <= w) { cout << i << endl; return 0; }





    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.4215440208742203x):
#include <iostream>

#include <cstring>

using namespace std;

const int maxn=1e5+5;

using ll = long long;

ll dp[maxn];

ll w[maxn],v[maxn];

int main()

{

   ll n,m,mm;

   cin>>n>>mm;

   m=0;

   memset(dp,0x3f,sizeof(dp));

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

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

       m+=v[i];

   }

   dp[0]=0;

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

       for(ll j=m;j>=0;--j){

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

          // cout<<dp[j]<<endl;

       }

   }

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

       if(dp[i]<=mm){

           cout<<i<<endl;

           break;

       }

   }

   return 0;

}
// Slower C++ solution:
//by Top1

#include <bits/stdc++.h>

     

#define f first

#define s second

#define int long long

#define ld long double

#define ull unsigned long long

#define pii pair <int, int>

#define pll pair <long, long>

#define mp make_pair

#define pb push_back

#define pf push_front

#define pp pop_back

#define rev reverse

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

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

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

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

#define nxtp next_permutation

#define pvtp prev_permutation

#define KZ return 

#define x1 x21212121

#define y1 y21212121

#define sqr(x) x * x 

 

const int inf = (1e9) + 1;

const int mod = 1e9 + 7;

const int maxn = 1e5 + 111;

     

using namespace std;



int q, n, w, a[maxn], b[maxn], dp[105][maxn], ans;



void solve(){    

	cin >> n >> w;



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

		cin >> a[i] >> b[i];



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

		for (int j = 1e5; j >= 0; j--)

			dp[i][j] = inf;



	dp[0][0] = 0;



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

		for (int j = 1e5; j >= 0; j--){

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

			if (j - b[i] >= 0)

				dp[i][j] = min(dp[i][j], dp[i - 1][j - b[i]] + a[i]);

			if (dp[i][j] <= w)

				ans = max(ans, j);

		}

		/*for (int j = 1; j <= 20; j++)

			cout << dp[i][j] << " ";

		cout << endl;	*/

		//cout << dp[i][100000] << endl;

	}



	//cout << dp[n][100000] << " ";



	/*for (int i = 0; i <= w; i++)

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

	*/

	cout << ans;

}



main(){

    boost;

	    

	q = 1;



	while (q--){

		solve();

	}

}

// Faster C++ solution for the same problem (measured speedup: 4.069566962863965x):
#include <iostream>

#include <cstring>

using namespace std;

const int maxn=1e5+5;

using ll = long long;

ll dp[maxn];

ll w[maxn],v[maxn];

int main()

{

   ll n,m,mm;

   cin>>n>>mm;

   m=0;

   memset(dp,0x3f,sizeof(dp));

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

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

       m+=v[i];

   }

   dp[0]=0;

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

       for(ll j=m;j>=0;--j){

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

          // cout<<dp[j]<<endl;

       }

   }

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

       if(dp[i]<=mm){

           cout<<i<<endl;

           break;

       }

   }

   return 0;

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



using namespace std;



#define loop(i, n) for(int i = 0;i < int(n);i++)

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

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

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

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

#define RALL(c) c.rbegin(), c.rend()

#define PI acos(-1)

#define pb push_back

#define mp make_pair

#define fr first

#define sc second

#define sfi1(v) scanf("%d",&v)

#define sfi2(v1, v2) scanf("%d %d",&v1,&v2)

#define sfi3(v1, v2, v3) scanf("%d %d %d",&v1,&v2,&v3)

#define sfll1(v) scanf("%I64d",&v);

#define sfll2(v1, v2) scanf("%I64d %I64d",&v1,&v2)

#define sfll3(v1, v2, v3) scanf("%I64d %I64d %I64d",&v1,&v2,&v3)

#define endl '\n'



typedef vector<int> vi;

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

typedef long long ll;

typedef pair<int, int> pii;



const int N = 105;

const int V = 1e5 + 5;

int dp[2][V];

int w[N], v[N];

int n, W;





int main() {



#ifndef ONLINE_JUDGE

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

    //freopen("out.in", "w", stdout);

#endif

    ios_base::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);



    cin >> n >> W;

    range(i, 1, V - 1)dp[n & 1][i] = W + 2;



    loop(i, n)cin >> w[i] >> v[i];



    rloop(i, n - 1) {

        loop(j, V) {

            dp[i & 1][j] = dp[i & 1 ^ 1][j];

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

        }

    }



    int ans = 0;

    rloop(i, V - 1) {

        if (dp[0][i] <= W) {

            ans = i;

            break;

        }

    }



    cout << ans << endl;



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4675485931502026x):
#include <iostream>

#include <cstring>

using namespace std;

const int maxn=1e5+5;

using ll = long long;

ll dp[maxn];

ll w[maxn],v[maxn];

int main()

{

   ll n,m,mm;

   cin>>n>>mm;

   m=0;

   memset(dp,0x3f,sizeof(dp));

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

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

       m+=v[i];

   }

   dp[0]=0;

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

       for(ll j=m;j>=0;--j){

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

          // cout<<dp[j]<<endl;

       }

   }

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

       if(dp[i]<=mm){

           cout<<i<<endl;

           break;

       }

   }

   return 0;

}
// Slower C++ solution:
// @author: Amirul Islam



#include <bits/stdc++.h>

using namespace std;

#define LLI long long int

#define PII pair <int, int>



const int mx = 1e5+5;

const LLI INF = 1e10;



LLI dp[mx], w, val, N, W;



LLI max(LLI a, LLI b) {

    return (a > b ? a : b);

}



LLI knapsack(int N, int W) {

    for (int i = 0; i < mx; i++) dp[i] = INF;

    dp[0] = 0;

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

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

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

            if (j >= val) dp[j] = min(dp[j], dp[j-val] + w);

        }

    }

    LLI ret = 0;

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

        if (dp[i] <= W) ret = max(ret, i);

    }

    return ret;

} 



int main() {

    scanf("%lld %lld", &N, &W);

    printf("%lld\n", knapsack(N, W));

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.435691762546516x):
#include <iostream>

#include <cstring>

using namespace std;

const int maxn=1e5+5;

using ll = long long;

ll dp[maxn];

ll w[maxn],v[maxn];

int main()

{

   ll n,m,mm;

   cin>>n>>mm;

   m=0;

   memset(dp,0x3f,sizeof(dp));

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

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

       m+=v[i];

   }

   dp[0]=0;

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

       for(ll j=m;j>=0;--j){

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

          // cout<<dp[j]<<endl;

       }

   }

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

       if(dp[i]<=mm){

           cout<<i<<endl;

           break;

       }

   }

   return 0;

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

using namespace std;

typedef long long ll;

const int inf=0x3f3f3f3f;

const int N=1005;

const int maxn=1e5+5;

ll a[N],v[N];

ll dp[maxn];

int main(){

	int n,k;

	cin>>n>>k;

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

		cin>>a[i]>>v[i];

	memset(dp,0x3f,sizeof(dp));

	dp[0]=0;

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

		for(int j=maxn-5;j>=v[i];--j)

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

	ll ans=0;

	for(int i=1;i<=maxn-5;++i)

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

	cout<<ans<<endl;

	return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 1.4454874247843081x):
#include <iostream>

#include <cstring>

using namespace std;

const int maxn=1e5+5;

using ll = long long;

ll dp[maxn];

ll w[maxn],v[maxn];

int main()

{

   ll n,m,mm;

   cin>>n>>mm;

   m=0;

   memset(dp,0x3f,sizeof(dp));

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

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

       m+=v[i];

   }

   dp[0]=0;

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

       for(ll j=m;j>=0;--j){

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

          // cout<<dp[j]<<endl;

       }

   }

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

       if(dp[i]<=mm){

           cout<<i<<endl;

           break;

       }

   }

   return 0;

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

using namespace std;

typedef long long int ll;

const int MAX=1e5+10;

ll dp[MAX];

int main (){

    int N,W;cin >> N >> W;

    memset(dp,63,sizeof dp);

    dp[0]=0;

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

        int w,val;

        cin >> w >> val;

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

            if(j-val>=0)

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

        }



    }

    int res=0;

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

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

    }

    cout<<res<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.6590358189170995x):
#include <bits/stdc++.h>



#define forn(i, n) for (int i = 0; i < int(n); ++i)

#define for1(i, n) for (int i = 1; i <= int(n); ++i)

#define fore(i, l, r) for (int i = int(l); i <= int(r); ++i)

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

#define pb push_back

#define eb emplace_back

#define ms(a, x) memset(a, x, sizeof(a))

#define F first

#define S second

#define endl '\n'

#define de(x) cout<<#x<<" = "<<(x)<<endl

#define de2(x,y) cout<<#x<<" = "<<(x) <<' '<< #y<<" = "<<y<<endl;



using namespace std;



typedef long long ll;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());



const int N=1e5+5;

ll dp[N];

int v[105],w[105];

int main() {



    ios::sync_with_stdio(false);

    cin.tie(nullptr);

	int N,W;

	cin>>N>>W;

	int sum=0;

	forn(i,N){

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

		sum+=v[i];

	}

	ms(dp,INF);

	dp[0]=0;

	forn(i,N){

		for(int j=sum;j>=v[i];j--){

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

		}

	}

	ll ans=0;

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

		if(dp[i]<=W){

			ans=i;

			break;

		}

	}

	cout<<ans;

	return 0;

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

using namespace std;



#define pb push_back

#define mk make_pair

#define fi first

#define se second

#define eb emplace_back



typedef long long ll;

typedef pair<int,int> ii;

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

const ll INF = 0x3f3f3f3f3f3fef;



const int V = 1e5+2;

const int N = 102;



ll dp[V];

ll p[N];

ll v[N];

ll n,w;

 

int main() {

    ios_base::sync_with_stdio(false);

    for(int i = 0; i < V; i++) dp[i] = INF;



    cin >> n >> w;

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

    

    dp[0] = 0;



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

        for(ll j = V-1; j-v[i] >= 0; j--)

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



    for(int i = V-1; i >= 0; i--)

        if(dp[i] <= w) { cout << i << endl; return 0; }





    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.4065908736873247x):
#include <bits/stdc++.h>



#define forn(i, n) for (int i = 0; i < int(n); ++i)

#define for1(i, n) for (int i = 1; i <= int(n); ++i)

#define fore(i, l, r) for (int i = int(l); i <= int(r); ++i)

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

#define pb push_back

#define eb emplace_back

#define ms(a, x) memset(a, x, sizeof(a))

#define F first

#define S second

#define endl '\n'

#define de(x) cout<<#x<<" = "<<(x)<<endl

#define de2(x,y) cout<<#x<<" = "<<(x) <<' '<< #y<<" = "<<y<<endl;



using namespace std;



typedef long long ll;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());



const int N=1e5+5;

ll dp[N];

int v[105],w[105];

int main() {



    ios::sync_with_stdio(false);

    cin.tie(nullptr);

	int N,W;

	cin>>N>>W;

	int sum=0;

	forn(i,N){

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

		sum+=v[i];

	}

	ms(dp,INF);

	dp[0]=0;

	forn(i,N){

		for(int j=sum;j>=v[i];j--){

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

		}

	}

	ll ans=0;

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

		if(dp[i]<=W){

			ans=i;

			break;

		}

	}

	cout<<ans;

	return 0;

}
// Slower C++ solution:
//by Top1

#include <bits/stdc++.h>

     

#define f first

#define s second

#define int long long

#define ld long double

#define ull unsigned long long

#define pii pair <int, int>

#define pll pair <long, long>

#define mp make_pair

#define pb push_back

#define pf push_front

#define pp pop_back

#define rev reverse

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

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

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

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

#define nxtp next_permutation

#define pvtp prev_permutation

#define KZ return 

#define x1 x21212121

#define y1 y21212121

#define sqr(x) x * x 

 

const int inf = (1e9) + 1;

const int mod = 1e9 + 7;

const int maxn = 1e5 + 111;

     

using namespace std;



int q, n, w, a[maxn], b[maxn], dp[105][maxn], ans;



void solve(){    

	cin >> n >> w;



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

		cin >> a[i] >> b[i];



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

		for (int j = 1e5; j >= 0; j--)

			dp[i][j] = inf;



	dp[0][0] = 0;



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

		for (int j = 1e5; j >= 0; j--){

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

			if (j - b[i] >= 0)

				dp[i][j] = min(dp[i][j], dp[i - 1][j - b[i]] + a[i]);

			if (dp[i][j] <= w)

				ans = max(ans, j);

		}

		/*for (int j = 1; j <= 20; j++)

			cout << dp[i][j] << " ";

		cout << endl;	*/

		//cout << dp[i][100000] << endl;

	}



	//cout << dp[n][100000] << " ";



	/*for (int i = 0; i <= w; i++)

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

	*/

	cout << ans;

}



main(){

    boost;

	    

	q = 1;



	while (q--){

		solve();

	}

}

// Faster C++ solution for the same problem (measured speedup: 4.026759400882725x):
#include <bits/stdc++.h>



#define forn(i, n) for (int i = 0; i < int(n); ++i)

#define for1(i, n) for (int i = 1; i <= int(n); ++i)

#define fore(i, l, r) for (int i = int(l); i <= int(r); ++i)

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

#define pb push_back

#define eb emplace_back

#define ms(a, x) memset(a, x, sizeof(a))

#define F first

#define S second

#define endl '\n'

#define de(x) cout<<#x<<" = "<<(x)<<endl

#define de2(x,y) cout<<#x<<" = "<<(x) <<' '<< #y<<" = "<<y<<endl;



using namespace std;



typedef long long ll;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());



const int N=1e5+5;

ll dp[N];

int v[105],w[105];

int main() {



    ios::sync_with_stdio(false);

    cin.tie(nullptr);

	int N,W;

	cin>>N>>W;

	int sum=0;

	forn(i,N){

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

		sum+=v[i];

	}

	ms(dp,INF);

	dp[0]=0;

	forn(i,N){

		for(int j=sum;j>=v[i];j--){

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

		}

	}

	ll ans=0;

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

		if(dp[i]<=W){

			ans=i;

			break;

		}

	}

	cout<<ans;

	return 0;

}