#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
long long a,b,c;
int main()
{scanf("%lld%lld%lld",&a,&b,&c);//不知为何脑抽的开了long longif(b==0){printf("Yes");return 0;}if(a/b>=c)printf("Yes");else printf("No");
}

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
long long n,c[1000005],sum,minn=10000000001;
struct t{long long a,b;
}tong[1000005];
bool cmp(t k,t q)
{ return k.b<q.b;
}
int main()
{scanf("%lld",&n);for(long long i=1;i<=n;i++){scanf("%lld%lld",&tong[i].a,&tong[i].b);}sort(tong+1,tong+1+n,cmp);for(long long i=1;i<=n;i++){ sum+=tong[i].a;c[i]=tong[i].b-sum;if(c[i]<minn)minn=c[i];//其实等价于minn=min(minn,c[i]),而且c[i]好像可以直接写成c的样子
    }printf("%lld",minn);
}

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
long long zbz,xbz,zbx,ans,n,m,a[1000005],b[1000005],xbx[100005],xh[100005],bxh[100005],xs[100005];
const long long mod=10007;
int main()
{   cin>>n>>m;for(int i=1;i<=n;i++){cin>>b[i];}for(int i=1;i<=n;i++){cin>>a[i];}for(int j=1;j<=3;j++)     {for(long long i=j;i<=n;i+=3){xbz=(b[i]*xh[a[i]]+mod)%mod;zbx=(i*bxh[a[i]]+mod)%mod;zbz=((i%mod)*(b[i]%mod)*(xs[a[i]]%mod)%mod+mod)%mod;ans=(ans+xbx[a[i]]-xbz+zbx-zbz+mod)%mod;//注意先算答案，再更新数组bxh[a[i]]=(bxh[a[i]]+b[i]+mod)%mod;//注意取模xs[a[i]]++;//当前a[i]种类的x出现的数量xh[a[i]]=(xh[a[i]]+i+mod)%mod;xbx[a[i]]=(xbx[a[i]]+(i%mod)*(b[i]%mod)%mod+mod)%mod;}memset(xs,0,sizeof(xs));//一定要记住memsetmemset(xh,0,sizeof(xh));memset(xbx,0,sizeof(xbx));memset(bxh,0,sizeof(bxh));zbz=0;zbx=0;xbz=0;}ans=(ans+mod)%mod;cout<<ans;
}