#include<cstdio>

 #include<cstdlib>

 #include<cmath>

 #include<cstring>

 #include<algorithm>

 #include<iostream>

 #include<vector>

 #include<map>

 #include<set>

 #include<queue>

 #include<string>

 #define inf 1000000000

 #define maxn 100000+5

 #define maxm 500+100

 #define eps 1e-10

 #define ll long long

 #define pa pair<int,int>

 #define for0(i,n) for(int i=0;i<=(n);i++)

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

 #define for2(i,x,y) for(int i=(x);i<=(y);i++)

 #define for3(i,x,y) for(int i=(x);i>=(y);i--)

 #define mod 1000000007

 using namespace std;

 inline int read()

 {

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

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

     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}

     return x*f;

 }

 int n,m,h,s[maxn],a[maxn],b[maxn],id[maxn];

 inline void add(int x,int y){for(;x<=n;x+=x&(-x))s[x]+=y;}

 inline int sum(int x){int t=;for(;x;x-=x&(-x))t+=s[x];return t;}

 inline bool cmp(int x,int y){return b[x]==b[y]?x<y:b[x]<b[y];}

 int main()

 {

     freopen("input","r",stdin);

     freopen("output.txt","w",stdout);

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

     for1(i,n)a[i]=read();

     for1(i,n)

     {

         int x=read();

         if(a[i]<=h)b[i]=(h-a[i])/x+;//b[i]表示该点最多可以被饮几次水

         if(b[i]>)add(i,);id[i]=i;

     }

     //可以证明，最后的答案一定是某个b[i]，所以我们按b[i]从小到大处理

     sort(id+,id+n+,cmp);

     int now=,pos=,ans=,st=n+;//now 表示当前的趟数

     //for1(i,n)cout<<i<<' '<<id[i]<<' '<<b[i]<<endl;

     for1(i,n)if(b[id[i]]){st=i;break;}

     for2(i,st,n)

     {

         while(now<=m)

         {

             int t=sum(n)-sum(pos);

             if(ans+t<b[id[i]])ans+=t,now++,pos=;//如果可以走完这一趟

             else break;

         }

         if(now>m)break;

         int l=pos,r=n,t=sum(pos);

         while(l<=r)

         {

             int mid=(l+r)>>;

             if(sum(mid)-t+ans>=b[id[i]])r=mid-;else l=mid+;

         }//二分到下一个点

         ans=b[id[i]];pos=l;

         for(;b[id[i]]==b[id[i+]];i++)add(id[i],-);//去除不能再次饮水的点

         add(id[i],-);

     }

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

     return ;

 }