1742

/*
动态规划
限制个数的背包问题啊

*/

// include file
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <ctime>

#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <bitset>

#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <list>
#include <functional>

using namespace std;

// typedef
typedef long long LL;
typedef unsigned long long ULL;

// 
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#define FORi(a,b,c) for(int i=(a);i<(b);i+=c)
#define FORj(a,b,c) for(int j=(a);j<(b);j+=c)
#define FORk(a,b,c) for(int k=(a);k<(b);k+=c)
#define FORp(a,b,c) for(int p=(a);p<(b);p+=c)
#define FORii(a,b,c) for(int ii=(a);ii<(b);ii+=c)
#define FORjj(a,b,c) for(int jj=(a);jj<(b);jj+=c)
#define FORkk(a,b,c) for(int kk=(a);kk<(b);kk+=c)

#define FF(i,a)    for(int i=0;i<(a);i++)
#define FFD(i,a)   for(int i=(a)-1;i>=0;i--)

#define Z(a) (a<<1)
#define Y(a) (a>>1)

const double eps = 1e-6;
const double INFf = 1e100;
const int INFi = 2000000000;
const LL INFll = (LL)1<<62;
const double Pi = acos(-1.0);

template<class T> inline T sqr(T a){return a*a;}
template<class T> inline T TMAX(T x,T y)
{
	if(x>y) return x;
	return y;
}
template<class T> inline T TMIN(T x,T y)
{
	if(x<y) return x;
	return y;
}
template<class T> inline void SWAP(T &x,T &y)
{
	T t = x;
	x = y;
	y = t;
}
template<class T> inline T MMAX(T x,T y,T z)
{
	return TMAX(TMAX(x,y),z);
}
template<class T> inline T MMIN(T x,T y,T z)
{
	return TMIN(TMIN(x,y),z);
}


// code begin
int V[110];
int C[110];
int N,M;
bool DP[100010];
int cnt[100010][2];
// 
// 
int main()
{
	read;
	write;
	int ct,ans;
	while(scanf("%d %d",&N,&M)==2)
	{
		if(N+M==0) break;
		FORi(1,N+1,1)
		{
			scanf("%d",V+i);
		}
		FORi(1,N+1,1)
		{
			scanf("%d",C+i);
		}
		FORi(0,M+1,1) DP[i] = 0;
		DP[0] = true;
		ans = 0;
		FORi(1,N+1,1)
		{
			//
			FORj(0,M+1,1)
			{
				if( j+V[i]>M ) break;
				if( DP[j] && !DP[j+V[i]] )
				{
					if(cnt[j][0]!=i)
					{
						cnt[j][0] = i;
						cnt[j][1] = 0;
					}
					if( cnt[j][1]+1<=C[i] )
					{
						ans++;
						DP[j+V[i]] = true;
						cnt[j+V[i]][1] = cnt[j][1]+1;
						cnt[j+V[i]][0] = cnt[j][0];
					}
				}
			}
			if(ans==M) break;
		}
		printf("%d\n",ans);
	}
	return 0;
}