stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include <bits/stdc++.h>                                           

 

#define int long long

//#pragma GCC optimize("Ofast")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")

 

 

#define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout);

#define forev(i, b, a) for(int i = (b); i >= (a); --i)

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

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

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

#define pb push_back

#define ppb pop_back

#define mp make_pair

#define s second

#define f first

 

 

using namespace std;

 

 

typedef pair < long long, long long > pll;    

typedef pair < int, int > pii;

typedef unsigned long long ull;         

typedef vector < pii > vpii;

typedef vector < int > vi;

typedef long double ldb; 

typedef long long ll;  

typedef double db;                         

 

 

const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, block = 555;

const pii base = mp(1171, 3307), Mod = mp(1e9 + 7, 1e9 + 9);

const int inf = 1e9, N = 1e5 + 48, mod = 1e9 + 7, maxn = 102;

const db eps = 1e-12, pi = 3.14159265359;

const ll INF = 1e18;





int n, m, w[maxn], c[maxn], dp[N], M = N - 12;



main () {

	cin >> n >> m;

	forn (i, 1, n)

		cin >> w[i] >> c[i];





	dp[0] = 0;

	forn (i, 1, M)

		dp[i] = INF;





	forn (i, 1, n)

		forev (j, M, c[i])

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



	forev (i, M, 0)

		if (dp[i] <= m)  {

			cout << i << endl;

			return 0;

		}

}                       

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring> 

#include<iostream>

#include<algorithm>

using namespace std;

const int M=1200000;

int n,m,w[M],v[M],sum;long long dp[M];

int main(){

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

	dp[0]=0;

	for(int i=1;i<=n;i++) scanf("%d%d",&w[i],&v[i]),sum+=v[i];

	for(int i=1;i<=sum;i++) dp[i]=1000000000ll*100ll+1ll;

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

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

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

	for(int i=sum;i>=0;i--) if(dp[i]<=m){

		printf("%d",i) ;break;

	}

	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define ll long long

using namespace std;

const ll inf=0x7F7F7F7F7F7F7F7F;

ll n,wz,w[110],v[110],dp[100010];

inline int read(){

     int x=0,f=1;char ch=getchar();

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

     while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}

     return x*f;

 }

int main ( )

{

	n=read();wz=read();

	for(int i=1;i<=n;i++) w[i]=read(),v[i]=read();

	memset(dp,inf,sizeof(dp));dp[0]=0;

	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]);

    for(int i=100000;i>=0;i--)

      if(dp[i]<=wz)

      {

      	cout<<i;

      	break;

	  }

	return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring> 

#include<iostream>

#include<algorithm>

using namespace std;

const int M=1200000;

int n,m,w[M],v[M],sum;long long dp[M];

int main(){

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

	dp[0]=0;

	for(int i=1;i<=n;i++) scanf("%d%d",&w[i],&v[i]),sum+=v[i];

	for(int i=1;i<=sum;i++) dp[i]=1000000000ll*100ll+1ll;

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

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

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

	for(int i=sum;i>=0;i--) if(dp[i]<=m){

		printf("%d",i) ;break;

	}

	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

const int MX = 200005;

long long dp[MX];  // dp[i]表示价值为i时的最小体积

int main()

{

	int n,w;

	cin >> n >> w;

	for(int i = 0; i < MX; ++i) dp[i] = 1001002003004005006ll;

	dp[0] = 0;

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

	{

		int a,b;

		cin >> a >> b;

		for (int j = MX-1; j >= b; --j) 

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

	}

	int ans = 0;

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

		if (dp[i] <= w) ans = i;

	cout << ans << endl;

	return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring> 

#include<iostream>

#include<algorithm>

using namespace std;

const int M=1200000;

int n,m,w[M],v[M],sum;long long dp[M];

int main(){

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

	dp[0]=0;

	for(int i=1;i<=n;i++) scanf("%d%d",&w[i],&v[i]),sum+=v[i];

	for(int i=1;i<=sum;i++) dp[i]=1000000000ll*100ll+1ll;

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

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

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

	for(int i=sum;i>=0;i--) if(dp[i]<=m){

		printf("%d",i) ;break;

	}

	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

#define ll long long

#define ull unsigned long long

#define mst(a, b) memset((a), (b), sizeof(a))

#define mp(a, b) make_pair(a, b)

#define pi acos(-1)

#define pii pair<int, int>

#define pb push_back

const int INF = 0x3f3f3f3f;

const double eps = 1e-10;

const int maxn = 1e5 + 7;

const int maxm = 5e5 + 10;

ll w[maxn], v[maxn];

ll dp[maxn];

ll ans = 0;

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

    int n, m;

    cin >> n >> m;

    mst(dp, 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 = 1e5; j >= v[i]; --j)

        {

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

            if (dp[j] <= m)

                ans = max(ans, (ll)j);

        }

    }

    cout << ans << endl;

    return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring> 

#include<iostream>

#include<algorithm>

using namespace std;

const int M=1200000;

int n,m,w[M],v[M],sum;long long dp[M];

int main(){

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

	dp[0]=0;

	for(int i=1;i<=n;i++) scanf("%d%d",&w[i],&v[i]),sum+=v[i];

	for(int i=1;i<=sum;i++) dp[i]=1000000000ll*100ll+1ll;

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

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

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

	for(int i=sum;i>=0;i--) if(dp[i]<=m){

		printf("%d",i) ;break;

	}

	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

    // sum=0;

    // init();

    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;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring> 

#include<iostream>

#include<algorithm>

using namespace std;

const int M=1200000;

int n,m,w[M],v[M],sum;long long dp[M];

int main(){

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

	dp[0]=0;

	for(int i=1;i<=n;i++) scanf("%d%d",&w[i],&v[i]),sum+=v[i];

	for(int i=1;i<=sum;i++) dp[i]=1000000000ll*100ll+1ll;

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

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

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

	for(int i=sum;i>=0;i--) if(dp[i]<=m){

		printf("%d",i) ;break;

	}

	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<algorithm>

#include<cstring>

using namespace std;

int v[105],w[105];

long long dp[100005];

const int inf =0x3f3f3f3f;

int main()

{

	int n;

	long long W,V=0;

	cin>>n>>W;

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

	dp[0]=0;

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

		cin>>w[i]>>v[i],V+=v[i];

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

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

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

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

	long long ans;

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

		if(dp[i]<=W)

		 {

		 	ans=i;

		 	break;

		 }

	cout<<ans<<endl;	

	//long long ans=-1;

//	for(int i=0;i<=W;i++)

//		ans=max(ans,dp[i]);		

//	cout<<ans<<endl;

	return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring> 

#include<iostream>

#include<algorithm>

using namespace std;

const int M=1200000;

int n,m,w[M],v[M],sum;long long dp[M];

int main(){

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

	dp[0]=0;

	for(int i=1;i<=n;i++) scanf("%d%d",&w[i],&v[i]),sum+=v[i];

	for(int i=1;i<=sum;i++) dp[i]=1000000000ll*100ll+1ll;

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

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

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

	for(int i=sum;i>=0;i--) if(dp[i]<=m){

		printf("%d",i) ;break;

	}

	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<cstdio>

using namespace std;

int f[100010],w[101],v[101];

int main ()

{

	int n,z,i,j;

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

	for(i=1;i<=n;++i) scanf("%d%d",&w[i],&v[i]);

	for(i=1;i<=100000;++i) f[i]=z+1;

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

	  for(j=100000;j>=v[i];--j) f[j]=min(f[j],f[j-v[i]]+w[i]);

	for(i=100000;i>=0;--i) 

	  if(f[i]<=z)

	  {

	  	printf("%d",i);

	  	break;

	  }

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring> 

#include<iostream>

#include<algorithm>

using namespace std;

const int M=1200000;

int n,m,w[M],v[M],sum;long long dp[M];

int main(){

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

	dp[0]=0;

	for(int i=1;i<=n;i++) scanf("%d%d",&w[i],&v[i]),sum+=v[i];

	for(int i=1;i<=sum;i++) dp[i]=1000000000ll*100ll+1ll;

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

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

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

	for(int i=sum;i>=0;i--) if(dp[i]<=m){

		printf("%d",i) ;break;

	}

	

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
//I

//01背包

#include<cstdio>

#include<iostream> 

#include<cstring>

using namespace std;

int w[110],c[110];

long long f[100010];//f[v]表示不超过v价值的最小体积 

int main()

{

	for(int i=0;i<100010;i++)

		f[i]=1000000010;

	long long m,n,flag=0; 

	scanf("%d%d",&n,&m);//背包容量m和物品数量n 

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

		scanf("%d%d",&w[i],&c[i]);//重量和价值 

	f[0]=0;

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

		for(int v=100010;v>=c[i];v--)

			if(f[v-c[i]]+w[i]<f[v])

			{

				f[v]=f[v-c[i]]+w[i];

				if(f[v]<=m&&v>flag)

					flag=v;

			}

	printf("%lld",flag);

 } 

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define ll long long

#define inf 9999999999999999

#define F first

#define S second

#define eb emplace_back

#define mpp make_pair

#define P(x) cout<<__LINE__ <<": "<<#x<<' '<<(x)<<'\n'

#define dbg cout<<__LINE__ <<": 0manush\n"

#define read(x) freopen("input.txt","r",stdin)

#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

#define pi 2*acos(0.0)

#define ee exp(1)

#define mx 100009

#define mod 10000007

using namespace std;

ll dp[109][mx];



int main(){//read(x);

//fast;

ll r,g,h,i,j,t,e,a,b,c,d,n,l,k,z=0;

cin>>n>>k;

ll w[n],p[n];l=0;

for(i=0;i<n;i++){cin>>w[i]>>p[i];l+=p[i];}

dp[n][0]=0;

for(i=1;i<=l;i++)dp[n][i]=inf;

for(i=n-1;i>=0;i--){

    for(j=0;j<=l;j++){

        dp[i][j]=min(dp[i+1][j],(j-p[i]>=0)?w[i]+dp[i+1][j-p[i]]:inf);

        //printf("%2lld ",dp[i][j]);

    }//cout<<endl;

}

a=0;

    for(j=0;j<=l;j++){

        if(dp[0][j]<=k){

            a=max(a,j);

//          cout<<j<<' '<<dp[0][j]<<endl;

        }

    }//cout<<endl;

    cout<<a<<endl;

}






// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

#define ll long long

ll dp[110000];

ll ww[110],v[110];

int main()

{

    int n,w;cin>>n>>w;

    for(int i=1;i<=n;i++)cin>>ww[i]>>v[i];

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

    dp[0]=0;

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

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

            if(dp[j-v[i]]+ww[i]<=w)

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

    int maxx=0;

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

        if(dp[i]<=w)maxx=i;

    cout<<maxx<<endl;

    return 0;

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define ll long long

#define inf 9999999999999999

#define F first

#define S second

#define eb emplace_back

#define mpp make_pair

#define P(x) cout<<__LINE__ <<": "<<#x<<' '<<(x)<<'\n'

#define dbg cout<<__LINE__ <<": 0manush\n"

#define read(x) freopen("input.txt","r",stdin)

#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

#define pi 2*acos(0.0)

#define ee exp(1)

#define mx 100009

#define mod 10000007

using namespace std;

ll dp[109][mx];



int main(){//read(x);

//fast;

ll r,g,h,i,j,t,e,a,b,c,d,n,l,k,z=0;

cin>>n>>k;

ll w[n],p[n];l=0;

for(i=0;i<n;i++){cin>>w[i]>>p[i];l+=p[i];}

dp[n][0]=0;

for(i=1;i<=l;i++)dp[n][i]=inf;

for(i=n-1;i>=0;i--){

    for(j=0;j<=l;j++){

        dp[i][j]=min(dp[i+1][j],(j-p[i]>=0)?w[i]+dp[i+1][j-p[i]]:inf);

        //printf("%2lld ",dp[i][j]);

    }//cout<<endl;

}

a=0;

    for(j=0;j<=l;j++){

        if(dp[0][j]<=k){

            a=max(a,j);

//          cout<<j<<' '<<dp[0][j]<<endl;

        }

    }//cout<<endl;

    cout<<a<<endl;

}






// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



using namespace std;



#define nl '\n'

#define F first

#define S second

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

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

#define pb push_back



typedef long long ll;

typedef pair<int, int> pii;



const int N = (int)2e5 + 123;

const int K = 500;

const int maxn = (int)2e9;

const int mod = 100000007;

const ll inf = 1ll * 1e18 + 123;



int n;

ll w;

pair<ll, int> p[111];

ll a[100111];

int main(){



   ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0);



    #ifdef LOCAL

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

    #endif



    cin >> n >> w;

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

        cin >> p[i].F >> p[i].S;

    }

    for(int i = 0; i <= 100011; ++i) a[i] = -1;

    int sum = 0;

    a[0] = 0;

    sort(p + 1, p + n + 1);

    int ans = 0;

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

        sum += p[i].S;

        for(int j = sum; j >= p[i].S; --j){

            if(a[j - p[i].S] != -1){

                ll newW = a[j - p[i].S] + p[i].F;

                if(newW <= w){

                    ans = max(ans, j);

                }

                if(a[j] == -1 || a[j] > newW){

                    a[j] = newW;

                }

            }

        }

    }

    cout << ans;

    return 0;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define ll long long

#define inf 9999999999999999

#define F first

#define S second

#define eb emplace_back

#define mpp make_pair

#define P(x) cout<<__LINE__ <<": "<<#x<<' '<<(x)<<'\n'

#define dbg cout<<__LINE__ <<": 0manush\n"

#define read(x) freopen("input.txt","r",stdin)

#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

#define pi 2*acos(0.0)

#define ee exp(1)

#define mx 100009

#define mod 10000007

using namespace std;

ll dp[109][mx];



int main(){//read(x);

//fast;

ll r,g,h,i,j,t,e,a,b,c,d,n,l,k,z=0;

cin>>n>>k;

ll w[n],p[n];l=0;

for(i=0;i<n;i++){cin>>w[i]>>p[i];l+=p[i];}

dp[n][0]=0;

for(i=1;i<=l;i++)dp[n][i]=inf;

for(i=n-1;i>=0;i--){

    for(j=0;j<=l;j++){

        dp[i][j]=min(dp[i+1][j],(j-p[i]>=0)?w[i]+dp[i+1][j-p[i]]:inf);

        //printf("%2lld ",dp[i][j]);

    }//cout<<endl;

}

a=0;

    for(j=0;j<=l;j++){

        if(dp[0][j]<=k){

            a=max(a,j);

//          cout<<j<<' '<<dp[0][j]<<endl;

        }

    }//cout<<endl;

    cout<<a<<endl;

}






// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

#define     FasterIO        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define     bug(a,b)        cout<<a<<"\t"<<b<<"\n";

#define     Case(a,b)       cout<<"Case "<<a<<": "<<b<<"\n";

#define     watch(a)        cout<<fixed<<setprecision(a)

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

#define     Endl            "\n";

#define     tab             "\t"

#define     reset(a,b)      memset(a,b,sizeof(a));

#define     sfl(a)          scanf(" %lld", &a);

#define     sf(a)           scanf(" %d", &a);

#define     srt(a)          sort(a.begin(),a.end());

#define     pi              2*acos(0.0)

const int sz=1e6+9;

const ll inf=1e18L+9;

ll dp[sz];





void solve(){

    reset(dp,inf);

    int n,w;

    sf(n);

    sf(w);

    vector<pair<ll,int>>v(n);

    dp[0]=0;

    int pnt=0;

    lp(i,0,n){

       sfl(v[i].first);

       sf(v[i].second);

       pnt+=v[i].second;

    }

    //bug(pnt, 1);

   // sort(v.begin(),v.end(),com);

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



        for(int j=pnt-v[i].second;j>=0;j--){

            dp[j+v[i].second]=min( dp[j+v[i].second],(ll)dp[j]+(ll)v[i].first);

        }



    }

   // bug("140 dp",dp[140]);

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

        if(dp[i]<=w){

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

            return;

        }

    }

}



int main(){

    //FasterIO;

//    #ifndef ONLINE_JUDGE

 //       freopen("1.txt","r",stdin);

//    #endif

    //int t;

//cin>>t;

   // while(t--)

        solve();

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define ll long long

#define inf 9999999999999999

#define F first

#define S second

#define eb emplace_back

#define mpp make_pair

#define P(x) cout<<__LINE__ <<": "<<#x<<' '<<(x)<<'\n'

#define dbg cout<<__LINE__ <<": 0manush\n"

#define read(x) freopen("input.txt","r",stdin)

#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

#define pi 2*acos(0.0)

#define ee exp(1)

#define mx 100009

#define mod 10000007

using namespace std;

ll dp[109][mx];



int main(){//read(x);

//fast;

ll r,g,h,i,j,t,e,a,b,c,d,n,l,k,z=0;

cin>>n>>k;

ll w[n],p[n];l=0;

for(i=0;i<n;i++){cin>>w[i]>>p[i];l+=p[i];}

dp[n][0]=0;

for(i=1;i<=l;i++)dp[n][i]=inf;

for(i=n-1;i>=0;i--){

    for(j=0;j<=l;j++){

        dp[i][j]=min(dp[i+1][j],(j-p[i]>=0)?w[i]+dp[i+1][j-p[i]]:inf);

        //printf("%2lld ",dp[i][j]);

    }//cout<<endl;

}

a=0;

    for(j=0;j<=l;j++){

        if(dp[0][j]<=k){

            a=max(a,j);

//          cout<<j<<' '<<dp[0][j]<<endl;

        }

    }//cout<<endl;

    cout<<a<<endl;

}






// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const ll max_n=2e5+100;

const ll inf=0x3f3f3f3f;

ll t,n,m,p[max_n],v[max_n],q[max_n],dp[max_n];

int main(){

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

    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);

    cin>>n>>m;

    ll wmax=0,sum=0;

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

        cin>>q[i]>>p[i];

        sum+=p[i];

    }

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

    ll ans=0;

    dp[0]=0;

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

        for(ll j=sum;j>=p[i];j--){

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

            if(dp[j]<=m){

                ans=max(ans,j);

            }

        }

    }cout<<ans<<endl;

    return 0;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define ll long long

#define inf 9999999999999999

#define F first

#define S second

#define eb emplace_back

#define mpp make_pair

#define P(x) cout<<__LINE__ <<": "<<#x<<' '<<(x)<<'\n'

#define dbg cout<<__LINE__ <<": 0manush\n"

#define read(x) freopen("input.txt","r",stdin)

#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

#define pi 2*acos(0.0)

#define ee exp(1)

#define mx 100009

#define mod 10000007

using namespace std;

ll dp[109][mx];



int main(){//read(x);

//fast;

ll r,g,h,i,j,t,e,a,b,c,d,n,l,k,z=0;

cin>>n>>k;

ll w[n],p[n];l=0;

for(i=0;i<n;i++){cin>>w[i]>>p[i];l+=p[i];}

dp[n][0]=0;

for(i=1;i<=l;i++)dp[n][i]=inf;

for(i=n-1;i>=0;i--){

    for(j=0;j<=l;j++){

        dp[i][j]=min(dp[i+1][j],(j-p[i]>=0)?w[i]+dp[i+1][j-p[i]]:inf);

        //printf("%2lld ",dp[i][j]);

    }//cout<<endl;

}

a=0;

    for(j=0;j<=l;j++){

        if(dp[0][j]<=k){

            a=max(a,j);

//          cout<<j<<' '<<dp[0][j]<<endl;

        }

    }//cout<<endl;

    cout<<a<<endl;

}






// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

//dp[j]==>价值为j时体积为dp[j]

//体积dp[j]<w时的最大价值 



ll c[100005];

ll ans;

ll n,w;

ll a[100005],b[1000005];

int main()

{

	ll i,j,k,x,y,t;

	cin>>n>>w;//w是容积 

	memset(c,1e9+5,sizeof(c));

	//cout<<"ans="<<c[5]<<endl;

	c[0]=0;

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

	cin>>a[i]>>b[i];//a是体积 b是价值 

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

	{

		for(j=1e5;j>=b[i];--j)//1e2*1e3

		{

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

			if(c[j]<=w)

		    ans=max(ans,j);		

		}	

	}	

	cout<<ans;

	return 0;

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define ll long long

#define inf 9999999999999999

#define F first

#define S second

#define eb emplace_back

#define mpp make_pair

#define P(x) cout<<__LINE__ <<": "<<#x<<' '<<(x)<<'\n'

#define dbg cout<<__LINE__ <<": 0manush\n"

#define read(x) freopen("input.txt","r",stdin)

#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

#define pi 2*acos(0.0)

#define ee exp(1)

#define mx 100009

#define mod 10000007

using namespace std;

ll dp[109][mx];



int main(){//read(x);

//fast;

ll r,g,h,i,j,t,e,a,b,c,d,n,l,k,z=0;

cin>>n>>k;

ll w[n],p[n];l=0;

for(i=0;i<n;i++){cin>>w[i]>>p[i];l+=p[i];}

dp[n][0]=0;

for(i=1;i<=l;i++)dp[n][i]=inf;

for(i=n-1;i>=0;i--){

    for(j=0;j<=l;j++){

        dp[i][j]=min(dp[i+1][j],(j-p[i]>=0)?w[i]+dp[i+1][j-p[i]]:inf);

        //printf("%2lld ",dp[i][j]);

    }//cout<<endl;

}

a=0;

    for(j=0;j<=l;j++){

        if(dp[0][j]<=k){

            a=max(a,j);

//          cout<<j<<' '<<dp[0][j]<<endl;

        }

    }//cout<<endl;

    cout<<a<<endl;

}






// Fastest known solution for this problem: