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

using namespace std;

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

#define ll long long

#define fastIO cout << fixed << setprecision(9), ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)

double const EPS = 1e-9, PI = acos(-1);

const int N = 2e5 + 9, M = 5e2 + 9, OO = 1e9 + 7, MOD = 1e9 + 9;

const ll inf = 1e18;



ll dp[N], weight[109];

int val[109];



int main() {

  fastIO;

#ifdef LOCAL

  freopen("input.in", "rt", stdin);

#endif

  int n, k;

  cin >> n >> k;

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

    cin >> weight[i] >> val[i];

  memset(dp, '?', sizeof dp);

  int ans = 0;

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

    dp[0] = 0;

//    dp[i][val[i]] = min(dp[i][val[i]], weight[i]);

    if(dp[val[i]] <= k)

      ans = max(ans, val[i]);

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

      if(j + val[i] >= N) continue;

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

      if(dp[j + val[i]] <= k)

        ans = max(ans, j + val[i]);

    }

  }

  cout << ans;

  return 0;

}


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

#include<cstdio>

#define maxn 110

using namespace std;

int n,W,sum;

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

int main(){

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

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

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

		sum+=v[i];

	}

	for(int i=1;i<=sum;i++)f[i]=1000000001;

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

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

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

		}

	}

	int ans=0;

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

		if(f[i]<=W){

			ans=i;

			break;

		}

	}

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

	return 0;

} 

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

#include<cmath>

#include<cstdio>

#include<cstring>

using namespace std;

long long n,m,f[100010],w[100200],v[100200];

int main ()

{

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

	f[0]=0;

	cin>>n>>m;

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

	{

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

	}

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

	{

		for(long long j=100000;j>=w[i];j--)

		{

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

		}

	}	

	long long ans=0;

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

	{

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

	}

	cout<<ans;

}


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

#include<cstdio>

#define maxn 110

using namespace std;

int n,W,sum;

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

int main(){

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

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

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

		sum+=v[i];

	}

	for(int i=1;i<=sum;i++)f[i]=1000000001;

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

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

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

		}

	}

	int ans=0;

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

		if(f[i]<=W){

			ans=i;

			break;

		}

	}

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

	return 0;

} 

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

using namespace std;

typedef long long ll;

const int N=120,V=2e5;



int w[N],v[N];

ll d[V];



int main(){

	ios::sync_with_stdio(0);

	int n,W,ans=0;

	cin>>n>>W;

	memset(d,63,sizeof d);

	d[0]=0;

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

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

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

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

			if(d[j]<=W)ans=max(ans,j);

		}

	}

	cout<<ans;

	return 0;

}

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

#include<cstdio>

#define maxn 110

using namespace std;

int n,W,sum;

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

int main(){

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

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

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

		sum+=v[i];

	}

	for(int i=1;i<=sum;i++)f[i]=1000000001;

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

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

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

		}

	}

	int ans=0;

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

		if(f[i]<=W){

			ans=i;

			break;

		}

	}

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

	return 0;

} 

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

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<cmath>

#define LL long long

using namespace std;

LL n,m,dp[1000001],v[101],w[101],sum,ans; 

int main()

{

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

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

	{

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

		sum+=v[i];

	}

	memset(dp,127/3,sizeof dp);

	dp[0]=0;

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

	{

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

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

	}

	for(int i=sum;;i--)

	{

		if(dp[i]<=m){

			ans=i;break;

		}

	}

	cout<<ans;

	return 0;

}

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

#include<cstdio>

#define maxn 110

using namespace std;

int n,W,sum;

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

int main(){

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

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

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

		sum+=v[i];

	}

	for(int i=1;i<=sum;i++)f[i]=1000000001;

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

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

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

		}

	}

	int ans=0;

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

		if(f[i]<=W){

			ans=i;

			break;

		}

	}

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

	return 0;

} 

// Fastest known solution for this problem:
// Original C++ solution (slower):
// Author : Mohamed Sameh

#include <bits/stdc++.h>

#define pb push_back

#define f first

#define s second

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

#define rall(v) v.rbegin(),v.rend()

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

#define Flush fflush(stdout);

typedef long long ll ;

using namespace std ;

const int M = 100 ;

const int MAXV = 1e5+1; 

const ll MAX = 2e18+9;

ll dp[M][MAXV];

int w[M],v[M];

int N , W ;

int ans ;

int main ()

{   

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

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

    {

        scanf ("%d%d" ,w+i , v+i); 

    }

    // dp[i][j] means 

    // what is the minumum weight to get the value j in range [0 : i]

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

        for (int j =0 ; j < MAXV ;j++)

            dp[i][j] = MAX ; 

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

        dp[i][ v[i] ] = w[i];

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

    {

        for (int j = 1 ; j < MAXV ;j++)

        {

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

            if (j >= v[i])

            {

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

            }

        }

    }

    

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

    {

        if (dp[N-1][i] <= W )

            ans = max (ans , i); 

    }

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



}

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

using namespace std;

int n,W;

//long long dp[101][100001];

long long dp[100001];

int weigth[101];

int value[101];



int solve2() {

    int total = 0;

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

        total += value[i];

    }

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

        dp[i] = 2e9;

    dp[0] = 0;

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

        for(int v = total;v >= value[i];--v)

            dp[v] = min(dp[v], dp[v - value[i]] + weigth[i]);

    int ans = 0;

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

        if(dp[i] <= W)

            ans = max(ans, i);

    return ans;

}

int main(int argc, char const *argv[]) {

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

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

        scanf("%d %d", &weigth[i], &value[i]);

    printf("%d\n", solve2());



    return 0;

}

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

#include<cstdio>

#include<cstring>

#define N 110

#define LL long long

using namespace std;

int n,m;

LL w[N],v[N],dp[100010];

int main(){

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

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

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

	memset(dp,128/3,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]);

	LL ans=0;

	for(int i=0;i<=100000;i++)

		if(dp[i]<=(LL)m) ans=(LL)i;

	cout<<ans;

	return 0;

}

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

using namespace std;

int n,W;

//long long dp[101][100001];

long long dp[100001];

int weigth[101];

int value[101];



int solve2() {

    int total = 0;

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

        total += value[i];

    }

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

        dp[i] = 2e9;

    dp[0] = 0;

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

        for(int v = total;v >= value[i];--v)

            dp[v] = min(dp[v], dp[v - value[i]] + weigth[i]);

    int ans = 0;

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

        if(dp[i] <= W)

            ans = max(ans, i);

    return ans;

}

int main(int argc, char const *argv[]) {

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

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

        scanf("%d %d", &weigth[i], &value[i]);

    printf("%d\n", solve2());



    return 0;

}

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

#include<bits/stdc++.h>

using namespace std;

#define clr(arr, x) memset(arr, x, sizeof arr)

#define endl "\n"

#define Accepted 0

typedef long long ll;

typedef vector<int> vi;

typedef vector<ll> vl;

const int N = 1e5 + 5;

ll dp[N];

int main()

{

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

	int n, W;

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

	vi val(n + 1), cap(n + 1);

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

		scanf("%d%d", &cap[i], &val[i]);

	dp[0] = 0;

	ll ans = 0;

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

		for (int j = N-1; j>=val[i]; j--) {

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

			if (dp[j] <= W)ans = max(ans, (ll)j);

		}

	}

	

	printf("%lld", ans);



	return Accepted;

}

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

using namespace std;

int n,W;

//long long dp[101][100001];

long long dp[100001];

int weigth[101];

int value[101];



int solve2() {

    int total = 0;

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

        total += value[i];

    }

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

        dp[i] = 2e9;

    dp[0] = 0;

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

        for(int v = total;v >= value[i];--v)

            dp[v] = min(dp[v], dp[v - value[i]] + weigth[i]);

    int ans = 0;

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

        if(dp[i] <= W)

            ans = max(ans, i);

    return ans;

}

int main(int argc, char const *argv[]) {

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

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

        scanf("%d %d", &weigth[i], &value[i]);

    printf("%d\n", solve2());



    return 0;

}

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

#define Keep_calmYou_are_still_learning ios_base::sync_with_stdio(false);cin.tie(NULL);//freopen("out.txt","w",stdout);

typedef long long ll;

typedef long double ld;

using namespace std;





int main()

{

    Keep_calmYou_are_still_learning;



    ll n , m;

    cin>>n>>m;

    ll w[n],v[n],ans=0;

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

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



   ll dp[(int) 1e5 + 5];

   for (ll i=0;i<1e5+5;i++)

    dp[i]= 1e10;



    dp[0]=0;



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

    {

        for (ll j=1e5-v[i];j>=0;j--)

        {

            if (dp[j]<m)

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

            if (dp[j+v[i]]<=m)

                ans=max(ans,j+v[i]);

        }

    }

    cout<<ans<<'\n';

    return 0;

}


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

using namespace std;

int n,W;

//long long dp[101][100001];

long long dp[100001];

int weigth[101];

int value[101];



int solve2() {

    int total = 0;

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

        total += value[i];

    }

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

        dp[i] = 2e9;

    dp[0] = 0;

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

        for(int v = total;v >= value[i];--v)

            dp[v] = min(dp[v], dp[v - value[i]] + weigth[i]);

    int ans = 0;

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

        if(dp[i] <= W)

            ans = max(ans, i);

    return ans;

}

int main(int argc, char const *argv[]) {

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

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

        scanf("%d %d", &weigth[i], &value[i]);

    printf("%d\n", solve2());



    return 0;

}

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



using namespace std;

typedef long long ll;



const int N = 107;

int w[N], v[N];

int n, W;



ll dp[2][100007];



ll infll;



int main() {

    memset(&infll, 0x3f, sizeof infll);

    cin >> n >> W;



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

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

    }



    int cur = 0;

    memset(dp, 0x3f, sizeof dp);

    dp[1-cur][0] = 0;

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

        for(int j = 0; j < 100007; ++j) {

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

            if(j - v[i] >= 0 and dp[1-cur][j - v[i]] < infll) dp[cur][j] = min(dp[cur][j], w[i] + dp[1-cur][j - v[i]]);

        }

        cur = 1 - cur;

    }



    int ans = 0;

    for(int j = 0; j < 100007; ++j) {

        if(dp[1-cur][j] <= W) ans = max(ans, j);

    }



    cout << ans << endl;

}

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

using namespace std;

int n,W;

//long long dp[101][100001];

long long dp[100001];

int weigth[101];

int value[101];



int solve2() {

    int total = 0;

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

        total += value[i];

    }

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

        dp[i] = 2e9;

    dp[0] = 0;

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

        for(int v = total;v >= value[i];--v)

            dp[v] = min(dp[v], dp[v - value[i]] + weigth[i]);

    int ans = 0;

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

        if(dp[i] <= W)

            ans = max(ans, i);

    return ans;

}

int main(int argc, char const *argv[]) {

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

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

        scanf("%d %d", &weigth[i], &value[i]);

    printf("%d\n", solve2());



    return 0;

}

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

#include<bits/stdc++.h>

#include<iostream>

#include<algorithm>

#include<iomanip>

#include<string>

#include<cmath>

#include<vector>

#include<set>

#include<queue>

#include<stack>

#include<fstream>

#include<map>

#include<bitset>

#include<unordered_map>

#include<cstring>

#include<cctype>

using namespace std;

#define  ll long long

#define  ull unsigned long long

#define  ul unsigned long

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

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

#define pb(s) push_back(s);

#define format(n) fixed<<setprecision(n)

#define finl "\n"

#define Mohammed_Atef_Hassan fast();

#define mod 1000000007

#define INF LLONG_MAX

#define PI acos(-1)

#define clr(dp,n) memset(dp,n,sizeof dp)

#define ex(n) return cout<<n<<finl,0;

int dx[] = { 0, 1, 0, -1, 1, 1, -1, -1 };

int dy[] = { 1, 0, -1, 0, 1, -1, 1, -1 };

ll fix_mod(ll x, ll y)

{

	return (y + x % y) % y;

}

void fast() {

	std::ios_base::sync_with_stdio(false);

	cin.tie(nullptr); cout.tie(nullptr);

}

int main()

{

	Mohammed_Atef_Hassan

		int n; ll w;

	cin >> n >> w;

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

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

	vector<ll>mem(1e5 + 10, LLONG_MAX);

	mem[0] = 0;

	set<int>st{ 0 };

	vector<int>vv;

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

	{

		vv.clear();

		for (set<int>::reverse_iterator j=st.rbegin();j!=st.rend();j++)

		{

			if (*j + v[i].second <= 1e5)

				mem[*j + v[i].second] = min(mem[*j + v[i].second], v[i].first+mem[*j]),vv.push_back(*j+v[i].second);

		}

		for (auto j : vv)st.insert(j);

	}

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

	{

		if(mem[i]<=w)ex(i);

	}

}

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

using namespace std;

int n,W;

//long long dp[101][100001];

long long dp[100001];

int weigth[101];

int value[101];



int solve2() {

    int total = 0;

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

        total += value[i];

    }

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

        dp[i] = 2e9;

    dp[0] = 0;

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

        for(int v = total;v >= value[i];--v)

            dp[v] = min(dp[v], dp[v - value[i]] + weigth[i]);

    int ans = 0;

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

        if(dp[i] <= W)

            ans = max(ans, i);

    return ans;

}

int main(int argc, char const *argv[]) {

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

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

        scanf("%d %d", &weigth[i], &value[i]);

    printf("%d\n", solve2());



    return 0;

}

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

using namespace std;

#define INF 10e10

#define LL long long int

LL dp[100001];

int main() {

    ios_base::sync_with_stdio(false);

    int n,w,valSum=0;

    cin >> n >> w;

    int wgt[n+1],val[n+1];

    for(int i=1; i<100001; i++) dp[i] = INF;

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

        cin >> wgt[i] >> val[i];

        valSum += val[i];

    }

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

        for(int j=valSum; j>=val[i]; j--) {

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

        }

    }

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

        if(dp[i]<=w) {

            cout << i << "\n";

            break;

        }

    }

}


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

using namespace std;

int n,W;

//long long dp[101][100001];

long long dp[100001];

int weigth[101];

int value[101];



int solve2() {

    int total = 0;

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

        total += value[i];

    }

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

        dp[i] = 2e9;

    dp[0] = 0;

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

        for(int v = total;v >= value[i];--v)

            dp[v] = min(dp[v], dp[v - value[i]] + weigth[i]);

    int ans = 0;

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

        if(dp[i] <= W)

            ans = max(ans, i);

    return ans;

}

int main(int argc, char const *argv[]) {

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

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

        scanf("%d %d", &weigth[i], &value[i]);

    printf("%d\n", solve2());



    return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

#include<bits/stdc++.h>  

#define endl "\n"

using namespace std;

	typedef long long ll;

	const int maxn=100050;

	const int inf=0x3f3f3f3f;

	ll w[105];

	ll v[105];

	ll dp[maxn];

	ll n,m;

	int main(){

		IOS

		cin>>n>>m;

		ll sum=0;

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

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

			sum+=v[i];

		}

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

		dp[0]=0;	

		for(int i=2;i<=n+1;i++)

			for(ll j=sum;j>=0;j--)

				if(j-v[i-1]>=0)

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

			

		ll ans=-1;

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

			if(dp[j]<=m){

				ans=j;

				break;

			}

		}

		cout<<ans<<endl;

	}

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

using namespace std;



typedef long long ll;

#define dbg(x) cout<<(#x)<<" is "<<x<<endl

ll dp[102][100009],wt[102],v[102];



ll knapsack(ll n,ll s,ll w)

{

    ll a,b;

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

        for(int j=0; j<=s; j++){

            if(i==0) dp[i][j]=-1;

            else if(j==0) dp[i][j]=0;

            else if(v[i]>j){

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

            }

            else

            {

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

                b= dp[i-1][j-v[i]]==-1? -1:wt[i]+dp[i-1][j-v[i]];

                if(a==-1 && b==-1)

                {

                    dp[i][j]=-1;

                }

                else if(a==-1)

                {

                    dp[i][j]=b;

                }

                else if(b==-1){

                    dp[i][j]=a;

                }

                else{

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

                }

            }

        }

        dp[0][0]=0;

    }

    

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

        if(dp[n][j]<=w && dp[n][j]!=-1) return j;

    }

}

int main()

{

   ll n,w,i,s=0;

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

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

        scanf("%lld %lld",&wt[i],&v[i]);    

        s+=v[i];

   }

 

   ll ans=knapsack(n,s,w);

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

   return 0;

}

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

#include<cstring>

using namespace std;

int main()

{

	long long n,V;

	long long w[105],v[105],dp[100025];

    memset(dp,0x3f3f3f3f,sizeof(dp));

    cin>>n>>V;

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

    {

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

    }

    long long ans=0;

    dp[0]=0;

    for(int i=1;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]<=V)

                    ans=max(ans,1ll*j);

            }

        }

    }

    cout<<ans<<endl;

    return 0;

}

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

using namespace std;



typedef long long ll;

#define dbg(x) cout<<(#x)<<" is "<<x<<endl

ll dp[102][100009],wt[102],v[102];



ll knapsack(ll n,ll s,ll w)

{

    ll a,b;

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

        for(int j=0; j<=s; j++){

            if(i==0) dp[i][j]=-1;

            else if(j==0) dp[i][j]=0;

            else if(v[i]>j){

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

            }

            else

            {

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

                b= dp[i-1][j-v[i]]==-1? -1:wt[i]+dp[i-1][j-v[i]];

                if(a==-1 && b==-1)

                {

                    dp[i][j]=-1;

                }

                else if(a==-1)

                {

                    dp[i][j]=b;

                }

                else if(b==-1){

                    dp[i][j]=a;

                }

                else{

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

                }

            }

        }

        dp[0][0]=0;

    }

    

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

        if(dp[n][j]<=w && dp[n][j]!=-1) return j;

    }

}

int main()

{

   ll n,w,i,s=0;

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

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

        scanf("%lld %lld",&wt[i],&v[i]);    

        s+=v[i];

   }

 

   ll ans=knapsack(n,s,w);

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

   return 0;

}

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

#include<cstring>

using namespace std;

const int maxn=1e5+10;

const int inf=0x3f3f3f;

int v[110];

long long w[110];

long long dp[maxn];

int main()

{

	int n,m;

	cin>>n>>m;

	int ans=0;

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

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

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

	dp[0]=0;

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

	{

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

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

	}

	for(int i=1;i<=1e5;++i)

	{

		if(dp[i]<=m)

		 ans=max(ans,i);

	}

	cout<<ans<<endl;

	return 0;

}

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

using namespace std;



typedef long long ll;

#define dbg(x) cout<<(#x)<<" is "<<x<<endl

ll dp[102][100009],wt[102],v[102];



ll knapsack(ll n,ll s,ll w)

{

    ll a,b;

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

        for(int j=0; j<=s; j++){

            if(i==0) dp[i][j]=-1;

            else if(j==0) dp[i][j]=0;

            else if(v[i]>j){

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

            }

            else

            {

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

                b= dp[i-1][j-v[i]]==-1? -1:wt[i]+dp[i-1][j-v[i]];

                if(a==-1 && b==-1)

                {

                    dp[i][j]=-1;

                }

                else if(a==-1)

                {

                    dp[i][j]=b;

                }

                else if(b==-1){

                    dp[i][j]=a;

                }

                else{

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

                }

            }

        }

        dp[0][0]=0;

    }

    

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

        if(dp[n][j]<=w && dp[n][j]!=-1) return j;

    }

}

int main()

{

   ll n,w,i,s=0;

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

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

        scanf("%lld %lld",&wt[i],&v[i]);    

        s+=v[i];

   }

 

   ll ans=knapsack(n,s,w);

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

   return 0;

}

// Fastest known solution for this problem: