#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
const int P = 1e9 + 7;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
int power(int x, int y) {
  if (y == 0) return 1;
  int tmp = power(x, y / 2);
  if (y % 2 == 0)
    return 1ll * tmp * tmp % P;
  else
    return 1ll * tmp * tmp % P * x % P;
}
void update(int &x, int y) {
  x += y;
  if (x >= P) x -= P;
}
int miu[MAXN], powk[MAXN];
vector<int> factors[MAXN];
int tot, prime[MAXN], f[MAXN];
void sieve(int n) {
  miu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (f[i] == 0) prime[++tot] = f[i] = i, miu[i] = P - 1;
    for (int j = 1; j <= tot && prime[j] <= f[i]; j++) {
      int tmp = prime[j] * i;
      if (tmp > n) break;
      if (prime[j] == f[i])
        miu[tmp] = 0;
      else
        miu[tmp] = (P - miu[i]) % P;
      f[tmp] = prime[j];
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j += i) factors[j].push_back(i);
}
int calc(int lim, int sum) {
  int l = 1, r = lim;
  chkmax(l, sum - lim);
  chkmin(r, sum - 1);
  if (l > r) return 0;
  int ans = 0;
  for (auto x : factors[sum])
    update(ans, 1ll * miu[x] * (P - (l - 1) / x + r / x) % P);
  return ans;
}
int main() {
  int n, k, ans = 0;
  read(n), read(k);
  sieve(2 * n), powk[0] = 1;
  for (int i = 1; i <= n; i++) powk[i] = 1ll * powk[i - 1] * k % P;
  for (int s = 2; s <= 2 * n - 2; s++)
    for (auto g : factors[s]) {
      int lim = (n - 1) / g, sum = s / g;
      if (lim <= 0 || sum <= 1) continue;
      update(ans, 1ll * calc(lim, sum) * powk[max(g, s - n)] % P);
    }
  writeln(1ll * ans * power(powk[n], P - 2) % P);
  return 0;
}
