//特殊数据下会被卡成N^2(系统的nth_element并不会，因为稳定的sort大概用到了3种方法以保持其稳定性)
//特殊数据下会被卡成N^2(系统的nth_element并不会，因为稳定的sort大概用到了3种方法以保持其稳定性）
//特殊数据下会被卡成N^2(系统的nth_element并不会，因为稳定的sort大概用到了3种方法以保持其稳定性）
#include<ctime>
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
const int maxn=1e6+5;
ll s,gd[maxn];
int n,m,b[maxn],k[maxn],ans;
inline ll read()
{
ll t=0,f=1; char ch=getchar();
while(ch<'0'||ch>'9'){ if(ch=='-') f=-1; ch=getchar();}
while(ch>='0'&&ch<='9') t=(t<<3)+(t<<1)+(ch^48),ch=getchar();
return f*t;
}
inline bool cmp(ll a,ll b){ return a>b; }
void sort(int l,int r,int k){
if(l>=r) return;
int x=rand()%(r-l+1)+l;
ll tmp=gd[x]; int nl=l,nr=r;
swap(gd[x],gd[l]);
while(nl^nr){
while(gd[nr]<=tmp&&nr>nl) --nr;
while(gd[nl]>=tmp&&nl<nr) ++nl;
swap(gd[nl],gd[nr]);
}
swap(gd[nl],gd[l]);
if(nl-l+1>k) sort(l,nl-1,k);
if(k-(nl-l+1)>0) sort(nl+1,r,k-(nl-l+1));
}
inline bool check(int ans){
for(register int i=1;i<=n;++i) gd[i]=1ll*k[i]*ans+b[i];
sort(1,n,m);
ll sum=0;
for(register int i=1;i<=m;++i){
if(gd[i]<=0) continue;
if(s-sum<=gd[i]) return 1;
sum+=gd[i];
}
return sum>=s;
}
int main()
{
//freopen("merchant3.in","r",stdin);
srand(time(0));
n=read(),m=read(),s=read();
for(register int i=1;i<=n;++i) k[i]=read(),b[i]=read();
int l=1,r=1000000000,mid,sum=0;
if(check(0)){ puts("0"); return 0;}
while(l<=r){
int mid=(l+r)>>1;
if(check(mid)) ans=mid,r=mid-1;
else l=mid+1;
}
cout<<ans;
}