*思路:高精+快速幂求逆元
#include<bits/stdc++.h>
#define ll unsigned long long
#define p "19260817"
#define P 19260817
using namespace std;
template <typename T> void write(T x) {
if (x < 0) x = -x, putchar('-');
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
write(x);
puts("");
}
string jianf(string s1,string s2)
{
string st="";
while(s1.size()<s2.size()) s1="00"+s1;
while(s1.size()>s2.size()) s2="0"+s2;
for(int i=s1.size()-1;i>=0;i--)
{
if(s1[i]<'0') s1[i-1]--,s1[i]+=10;
int a=s1[i]-'0',b=s2[i]-'0';
if(a<b) s1[i-1]--,a+=10;
a-=b;
st=char(a+'0')+st;
}
while(st[0]=='0') st.erase(0,1);
return st;
}
bool cmp(string s1,string s2)
{
if(s1.size()>s2.size()) return 1;
if(s2.size()>s1.size()) return 0;
if(s1>=s2) return 1;
return 0;
}
string cf(string s1,string s2)
{
string s3=s1,st="";
s3=s1.substr(0,s2.size());
if(s3<s2) s3=s3+s1[s2.size()];
s1.erase(0,s3.size());
while(1)
{
int a=0;
while(cmp(s3,s2)){s3=jianf(s3,s2);a++;}
st=st+char(a+'0');
if(s1=="" )break;
while(!cmp(s3,s2)) s3=s3+s1[0],s1.erase(0,1);
}
if(s3=="") return "0";
return s3;
}
int niyuan(ll x)
{
ll k=x,p1=P-2,s=1;
while(p1!=0)
{
k%=P;
if(p1%2==1) s*=k;
s%=P;
p1=p1>>1;
k=k*k;
}
return s;
}
int main()
{
string s1="",s2="";
char ch=getchar();
for(;!isdigit(ch);ch=getchar()) ;
for(;isdigit(ch);ch=getchar()) s1=s1+ch;
for(;!isdigit(ch);ch=getchar()) ;
for(;isdigit(ch);ch=getchar()) s2=s2+ch;
s1=cf(s1,p);
s2=cf(s2,p);
ll a,b;
a=b=0;
for(int i=0;i<s1.size();i++) a=a*10+s1[i]-'0';
for(int i=0;i<s2.size();i++) b=b*10+s2[i]-'0';
if(b==0)
{
puts("Angry!");
return 0;
}
writeln((a*niyuan(b))%P);
return 0;
}