Problem: You are given integers n, k. Let's consider the alphabet consisting of k different elements.

Let beauty f(s) of the string s be the number of indexes i, 1≤ i<|s|, for which prefix of s of length i equals to suffix of s of length i. For example, beauty of the string abacaba equals 2, as for i = 1, 3 prefix and suffix of length i are equal.

Consider all words of length n in the given alphabet. Find the expected value of f(s)^2 of a uniformly chosen at random word. We can show that it can be expressed as P/Q, where P and Q are coprime and Q isn't divided by 10^9 + 7. Output P⋅ Q^{-1} mod 10^9 + 7.

Input

The first and the only line contains two integers n, k (1≤ n ≤ 10^5, 1≤ k≤ 10^9) — the length of a string and the size of alphabet respectively.

Output

Output a single integer — P× Q^{-1} mod 10^9 + 7.

Examples

Input


2 3


Output


333333336


Input


1 5


Output


0


Input


100 1


Output


9801


Input


10 10


Output


412377396

Note

In the first example, there are 9 words of length 2 in alphabet of size 3 — aa, ab, ac, ba, bb, bc, ca, cb, cc. 3 of them have beauty 1 and 6 of them have beauty 0, so the average value is 1/3.

In the third example, there is only one such word, and it has beauty 99, so the average value is 99^2.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, c, pw[200100], iv[200100], mu[200100], p[200100], b[200100];
int qpow(int a, int i) {
  int r = 1;
  for (; i; i >>= 1, a = 1ll * a * a % 1000000007)
    if (i & 1) r = 1ll * r * a % 1000000007;
  return r;
}
int s(int x, int t) {
  if (pw[x] == 1) return t;
  return 1ll * iv[x] * (pw[x * t] - 1) % 1000000007;
}
int S(int x, int t) {
  if (pw[x] == 1) return 1ll * t * (t + 1) / 2 % 1000000007;
  return -1ll * (1 + s(x, t - 1) - 1ll * pw[x * t] * t) % 1000000007 * iv[x] %
         1000000007;
}
int main() {
  scanf("%d %d", &n, &c);
  if (c == 1) {
    printf("%d", 1ll * (n - 1) * (n - 1) % 1000000007);
    return 0;
  }
  mu[1] = 1;
  for (int i = 2; i < 200100; ++i) {
    if (!b[i]) p[++p[0]] = i, mu[i] = -1;
    for (int j = 1; j <= p[0] && i * p[j] < 200100; ++j) {
      int k = i * p[j];
      mu[k] = -mu[i];
      b[k] = 1;
      if (i % p[j] == 0) {
        mu[k] = 0;
        break;
      }
    }
  }
  pw[0] = 1;
  for (int i = 1; i <= 200100; ++i)
    pw[i] = 1ll * c * pw[i - 1] % 1000000007,
    iv[i] = 1ll * qpow(pw[i] - 1, 1000000007 - 2) * pw[i] % 1000000007;
  int f, g, a, b;
  f = 0;
  for (int d = 1; d < n; ++d) {
    int t = 0;
    for (int k = 1, __ = (n - 1) / d; k <= __; ++k)
      b = (n - 1 + d) / d / k,
      t = (t + 1ll * mu[k] * b * (b - 1) / 2) % 1000000007;
    f = (f + 1ll * t * qpow(pw[n - d], 1000000007 - 2)) % 1000000007;
  }
  g = 0;
  for (int d = 1; d < n; ++d) {
    int t = 0;
    for (int k = 1, __ = (n - 1) / d; k <= __; ++k) {
      a = (n - 1) / d / k;
      b = (n - 1 + d) / d / k;
      t = (t + 1ll * mu[k] * (a + a + 1) * (s(d * k, a + a) - s(d * k, b))) %
          1000000007;
      t = (t - 1ll * mu[k] * (S(d * k, a + a) - S(d * k, b))) % 1000000007;
    }
    g = (g + 1ll * t * qpow(pw[n + n], 1000000007 - 2)) % 1000000007;
  }
  printf("%d", (1ll * f + g + 1000000007 + 1000000007) % 1000000007);
}