HDU 1452 欧拉定理

让你求$2004^x$所有因子之和，因子之和函数是积性函数$\sigma(n)=\sum_{d|n}d=\prod_{i=0}^{m}(\sum_{j=0}^{k_i}{P_i^{j}})$可用二项式定理证明，然后2004是给定的固定数，然后该怎么求就怎么求

/** @Date    : 2017-09-08 18:56:21
  * @FileName: HDU 1452 欧拉定理.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;
const LL mod = 29;

LL fpow(LL a, LL n)
{
	LL res = 1;
	while(n)
	{
		if(n & 1)
			res = a * res % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}
int main()
{
	//SUM factor = Sum(1->s)Sum(0->k)P[i]^k)  各个素因子各次和的乘积
	LL n;
	while(cin >> n && n)
	{
		LL INV2 = fpow(2, 27);
		LL INV166 = fpow(166, 27);
		LL ans = (((fpow(3, n + 1)-1) * INV2 % mod) * ((fpow(167, n + 1)-1) * INV166 % mod) * (fpow(2, 2 * n + 1)-1)) % mod; 
		printf("%lld\n", ans);
	}
    return 0;
}