|
|
#include <cstdio>
|
|
|
#include <algorithm>
|
|
|
#include <cstring>
|
|
|
using namespace std;
|
|
|
|
|
|
typedef long long LL;
|
|
|
const int maxn = 10000;
|
|
|
const LL Six = 166666668; /// 6，2关于mod的乘法逆元
|
|
|
const LL Two = 500000004;
|
|
|
const LL mod = 1e9 + 7; /// 尽量这样定义mod ，减少非必要的麻烦
|
|
|
|
|
|
bool book[maxn];
|
|
|
int pri[1300], cnt; /// 素数表
|
|
|
int factor[130]; /// 因子表
|
|
|
|
|
|
void Prime() {
|
|
|
memset(book, 0, sizeof(book));
|
|
|
cnt = 0;
|
|
|
book[0] = book[1] = 1;
|
|
|
for (int i = 2; i < maxn; i++) {
|
|
|
if (!book[i])
|
|
|
pri[cnt++] = i;
|
|
|
for (int j = 0; j < cnt && pri[j] * i < maxn; j++) {
|
|
|
book[i * pri[j]] = 1;
|
|
|
if (i % pri[j] == 0)
|
|
|
break;
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
|
|
|
inline LL Mod(LL a, LL b) {
|
|
|
return (a % mod) * (b % mod) % mod;
|
|
|
}
|
|
|
inline LL F(LL k, LL n) { /// 求(k*n)^2+k*n
|
|
|
return (Mod(k, k) * Mod(Mod(n, n + 1), Mod(n + n + 1, Six)) % mod + Mod(Mod(1 + n, n), Mod(k, Two))) % mod;
|
|
|
}
|
|
|
int main() {
|
|
|
LL n, m;
|
|
|
|
|
|
Prime();
|
|
|
while (~scanf("%lld%lld", &n, &m)) {
|
|
|
int fac_cnt = 0; /// 素数因子的个数
|
|
|
|
|
|
for (int i = 0; i < cnt; i++) {
|
|
|
if (pri[i] > m) break;
|
|
|
if (m % pri[i] == 0) {
|
|
|
factor[fac_cnt++] = pri[i];
|
|
|
while (m % pri[i] == 0) m /= pri[i];
|
|
|
}
|
|
|
}
|
|
|
if (m > 1) factor[fac_cnt++] = m;
|
|
|
|
|
|
LL sum = F(1LL, n), ans = 0; /// 计算总和sum
|
|
|
|
|
|
LL item = 1 << fac_cnt; /// 开始容斥
|
|
|
|
|
|
/// 例如有3个因子，那么item=1<<3=8(1000二进制)
|
|
|
/// 然后i从1开始枚举直到7(111二进制），i中二进制的位置1表式取这个位置的因子
|
|
|
/// 例如i=3(11二进制) 表示去前两个因子，i=5（101）表示取第1个和第3个的因子
|
|
|
for (int i = 1; i < item; i++) {
|
|
|
int num = 0, x = 1;
|
|
|
|
|
|
for (int j = 0; j < fac_cnt; j++) {
|
|
|
if (1 & (i >> j)) num++, x *= factor[j];
|
|
|
}
|
|
|
|
|
|
if (num & 1)
|
|
|
ans = (ans + F(x, n / x)) % mod; /// 根据容斥，取奇数个因子时，应加上
|
|
|
else
|
|
|
ans = ((ans - F(x, n / x)) % mod + mod) % mod;
|
|
|
}
|
|
|
|
|
|
printf("%lld\n", ((sum - ans) % mod + mod) % mod);
|
|
|
}
|
|
|
return 0;
|
|
|
}
|
