stablegradients/pie-gem5-pairs huggingface dataset:
// 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: 1.345553956888972x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<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: 3.8520241718147608x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    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: 1.3890993085291305x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    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: 1.3589454168147035x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    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: 1.3682174420850173x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    return 0;

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

#include <algorithm>

#include <cmath>

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <iomanip>

#include <string>

#include <set>

#include <vector>

#include <queue>

#include <stack>

using namespace std;

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

#define fore(i, a, n) for (long long i = a; i < n; i++)

#define ford(i, a, n) for (long long i = n - 1; i >= a; i--)

#define si(a) scanf("%d", &a)

#define sii(a, b) scanf("%d%d", &a, &b)

#define debug(a) cout << a << endl

#define endl '\n'

#define pi acos(-1.0)

#define tr t[root]

#define lson t[root << 1]

#define rson t[root << 1 | 1]

#define IO ios::sync_with_stdio(false), cin.tie(0)

const double eps = 1e-8;

inline int sgn(const double &x) { return x < -eps ? -1 : x > eps; }

typedef long long ll;

const int inf = 0x3f3f3f3f;

const int maxn=1000100;

#define maxn0 1e9+7;

//=================================================



ll sum;

ll w[maxn];

ll dp[maxn];

ll v[maxn];



// void init() {

//     for(ll i=0;i<=n;i++){

//         w[i]=0;

//         v[i]=0;

//     }

// }

int main() {

    IO;

    ll n,W;

    cin>>n>>W;

    memset(dp,inf,sizeof(dp));

    dp[0]=0;

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

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

        // sum+=v[i];

    }

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

        for(int j=100000;j>=v[i];j--) {

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

        }

    }

    ll ans=-1;

    for(int i=1;i<=100000;++i) {

        if(dp[i]<=W)

            ans=max(ans,1LL*i);

    }

    cout<<ans<<endl;



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.555249885149198x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    return 0;

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

using namespace std;

const int N = 1e5+100;

const int M = 200;

typedef long long ll;

int a[N],b[N],n,m;

ll f[N];



int main(){

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

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

		scanf("%d%d",&a[i],&b[i]);

	memset(f, 0x3f, sizeof f);

	f[0] = 0;

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

		for (int j = N - 10; j >= 0; j--)

			if (j - b[i] >= 0) f[j] = min(f[j], f[j-b[i]] + a[i]);



	int ans = 0;

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

		if (f[i] <= m) ans = i;

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.614929998380595x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    return 0;

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

#include<cstdlib>

#include<cstdio>

#include<algorithm>

#include<cmath>

#include<cstring>

#include<stack>

#define LL long long

using namespace std;

LL n,m;

LL dp[100011];

LL x,y,ans;

LL read()

{

	LL _=0,__=1;char ch=getchar();

	while(ch<'0'||ch>'9') {if(ch=='-')__=-1;ch=getchar();}

	while(ch>='0'&&ch<='9'){_=_*10+ch-'0';ch=getchar();}

	return _*__;

}

int main()

{

	n=read();m=read(); 

	for(int i=1;i<=100001;i++)

	dp[i]=1e9+7;

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

	{

		x=read();y=read();

		for(int j=100000;j>=y;j--)

		{

			dp[j]=min(dp[j],dp[j-y]+x);

		}

	}

	for(int i=1;i<=100001;i++)

	{

		if(dp[i]<=m)

		ans=max(ans,(LL)i);

	}

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

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.3435866121324611x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    return 0;

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

#define MAX 100005

#define pb push_back

#define mp make_pair

#define pii pair<long long,long long>

#define read(a) scanf("%d",&a)

#define p1(a) cout<<"Check "<<a<<endl;

#define p2(a,b) cout<<"Check "<<a<<' '<<b<<endl;

#define p3(a,b,c) cout<<"Check "<<a<<' '<<b<<' '<<c<<endl;

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

#define ll long long

#define clr(a) memset(a,0,sizeof(a))

#define MOD 10000000000

#define INF 10000000000000



using namespace std;



int n,cp;

ll cost[MAX];

ll weight[MAX];

ll dp[MAX];





int main()

{

    cin>>n>>cp;

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

        scanf("%lld%lld",&weight[i],&cost[i]);

    }

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

    dp[0]=0;

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

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

            if(j-cost[i]>=0 && dp[j-cost[i]]!=INF){

                dp[j]=min(dp[j],dp[j-cost[i]]+weight[i]);

            }

        }

    }

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

        if(dp[j]<=cp){

            cout<<j<<endl;

            break;

            //break;

        }

    }



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.272277085446725x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    return 0;

}
// 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: 1.184288623490119x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    return 0;

}
// Slower C++ solution:
#pragma region

#include <algorithm>

#include <cstring>

#include <iostream>

#include <map>

#include <queue>

#include <stack>

#include <string>

#include <vector>

typedef long long ll;

#define tr t[root]

#define lson t[root << 1]

#define rson t[root << 1 | 1]

#define rep(i, a, n) for (long long i = a; i <= n; ++i)

#define per(i, a, n) for (long long i = n; i >= a; --i)

#define IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)

template <typename T>

void read(T &x) {

    x = 0;

    char ch = getchar();

    ll f = 1;

    while (!isdigit(ch)) {

        if (ch == '-') f *= -1;

        ch = getchar();

    }

    while (isdigit(ch)) {

        x = x * 10 + ch - 48;

        ch = getchar();

    }

    x *= f;

}

using namespace std;

/*

 *                                                     __----~~~~~~~~~~~------___

 *                                    .  .   ~~//====......          __--~ ~~

 *                    -.            \_|//     |||\\  ~~~~~~::::... /~

 *                 ___-==_       _-~o~  \/    |||  \\            _/~~-

 *         __---~~~.==~||\=_    -_--~/_-~|-   |\\   \\        _/~

 *     _-~~     .=~    |  \\-_    '-~7  /-   /  ||    \      /

 *   .~       .~       |   \\ -_    /  /-   /   ||      \   /

 *  /  ____  /         |     \\ ~-_/  /|- _/   .||       \ /

 *  |~~    ~~|--~~~~--_ \     ~==-/   | \~--===~~        .\

 *           '         ~-|      /|    |-~\~~       __--~~

 *                       |-~~-_/ |    |   ~\_   _-~            /\

 *                            /  \     \__   \/~                \__

 *                        _--~ _/ | .-~~____--~-/                  ~~==.

 *                       ((->/~   '.|||' -_|    ~~-/ ,              . _||

 *                                  -_     ~\      ~~---l__i__i__i--~~_/

 *                                  _-~-__   ~)  \--______________--~~

 *                                //.-~~~-~_--~- |-------~~~~~~~~

 *                                       //.-~~~--\

 *                       ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 */



#pragma endregion

const int maxn = 1e5 + 5;

ll n, W, w[maxn], val[maxn], dp[maxn];

int main() {

    IO;

    cin >> n >> W;

    rep(i, 1, n) cin >> w[i] >> val[i];

    rep(i, 1, 1e5) dp[i] = 0x3f3f3f3f;

    dp[0] = 0;

    rep(i, 1, n) for (int t = 1e5; t >= val[i]; --t) dp[t] = min(dp[t], dp[t - val[i]] + w[i]);

    for (int i = 1e5; i; --i)

        if (dp[i] <= W) {

            cout << i << endl;

            return 0;

        }

}

// Faster C++ solution for the same problem (measured speedup: 1.34447038228186x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    return 0;

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

#define ll long long

using namespace std;

#define inf 0x3f3f3f3f 

#define MAXN 1019191130

int main(){

	ll  n , W; cin >> n >> W;

    ll w[101] , v[101] , dp[100001];

    for(int i = 0 ; i <= 100000; ++i) dp[i] = inf;

    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 =  100000 ; j >= v[i] ; j--){

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

        }

    }

    int j = 100001;

    while(j-- && dp[j] > W);



    cout<< j <<endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.3682372008792496x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    return 0;

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

using namespace std;

int main()

{

	 int n,W;

    cin>>n>>W;

    long long w[n];

    long long v[n];

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

    long long dp[100005];

    for(int i=1;i<100005;i++)

    {

    	dp[i]=1e15;

	}

    dp[0]=0;

    int res=0;

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

        for(int j=1e5;j>=v[i];j--){

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

            if(dp[j]<=W){

                res=max(res,j);

            }

        }

    }

    cout<<res<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.5159133648430896x):
#include <iostream>

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    return 0;

}