/*

 费马小定理

 题意：求s1+s2+s3+...+sn;si表示n划分i个数的n的划分的个数,如n=4,则s1=1,s2=3

     利用隔板定理可知，就是求（2^n-1）%mod-----Y

 */

 #include<stdio.h>

 #include<string.h>

 #include<stdlib.h>

 #include<algorithm>

 #include<iostream>

 #include<queue>

 #include<map>

 #include<stack>

 #include<set>

 #include<math.h>

 using namespace std;

 typedef long long int64;

 //typedef __int64 int64;

 typedef pair<int64,int64> PII;

 #define MP(a,b) make_pair((a),(b))

 const int maxn = ;

 const int inf = 0x7fffffff;

 const double pi=acos(-1.0);

 const double eps = 1e-;

 const int64 mod = +;

 int64 Fast_Pow( int64 a,int64 n,int64 mod ){

     int64 res = ;

     while( n>= ){

         if( n& ){

             res = res*a%mod;

         }

         a = a*a%mod;

         n >>= ;

     }

     return res%mod;

 }

 int64 GetNum( char str[],int64 mod ){

     int64 res = ;

     int len = strlen( str );

     for( int i=;i<len;i++ ){

         res = (res*+str[i]-'')%mod;

     }

     return res;

 }

 int main(){

     char str[ maxn ];

     while( scanf("%s",str)!=EOF ){

         int64 n = GetNum( str,mod- );

         printf("%I64d\n",Fast_Pow( ,(n-+mod)%mod,mod ));

     }

     return ;

 }