Happy 2004

Problem Description

Consider a positive integer X,and let S be the sum of all positive integer divisors of 2004^X. Your job is to determine S modulo 29 (the rest of the division of S by 29).

Take X = 1 for an example. The positive integer divisors of 2004^1 are 1, 2, 3, 4, 6, 12, 167, 334, 501, 668, 1002 and 2004. Therefore S = 4704 and S modulo 29 is equal to 6.

 

Input

The input consists of several test cases. Each test case contains a line with the integer X (1 <= X <= 10000000). 

A test case of X = 0 indicates the end of input, and should not be processed.

 

Output

For each test case, in a separate line, please output the result of S modulo 29.

 

Sample Input

1100000

 

Sample Output

610

2004有三个素因子2,3,167，（还可以看做是一个排列组合题目）

如图所示2004的素因子其实也就是2004的x次方的素因子，对于2004的x次方，ans=(2^0+2^1+2^2.......+2^e1)*(3^0+3^1+3^2+.....+3^e2)*(167^0+167^1+167^2+.......+167^e3)

每个小括号里是每个素因子可能形成的组合，乘起来就是所求的结果（理解不了的试着拆开写写），剩下就是合理运用取余，进行等比数列求和（注意这里要用到取模的逆元或者用二分等比数列求和）

#include<stdio.h>#define M 29#define LL long longLL multi(LL a,LL b){LL res=0;while(b){if(b&1){res=(res+a)%M;}b>>=1;a=(a+a)%M;}return res;}LL pow(LL a,LL b){LL res=1;while(b){if(b&1){res=multi(res,a);}b>>=1;a=multi(a,a);}return res;}int main(){LL x;while(scanf("%lld",&x),x){LL div_2,div_3,div_167;div_2=(pow(2,2*x+1)-1)%M;div_3=(pow(3,x+1)-1)*15%M;div_167=(pow(22,x+1)-1)*18%M;printf("%lld\n",(div_2*div_3*div_167)%M);}return 0;}