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