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;
const int mod = 1000000007;
long long powM(long long a, int t = mod - 2) {
  long long ret = 1;
  while (t) {
    if (t & 1) ret = ret * a % mod;
    a = a * a % mod;
    t >>= 1;
  }
  return ret;
}
int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }
bool e[100500];
int mu[100500], p[100500 >> 2], tn;
vector<int> sd[100500];
void sieve(int n) {
  mu[1] = e[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!e[i]) mu[p[++tn] = i] = -1;
    for (int j = 1; j <= tn && i * p[j] <= n; j++) {
      e[i * p[j]] = 1;
      if (i % p[j] == 0) break;
      mu[i * p[j]] = -mu[i];
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j += i) sd[j].push_back(i);
}
int n, c, pw[100500 << 1];
int calc1() {
  int ret = 0;
  for (int k = 1; k <= 2 * n - 2; k++)
    ret = (ret + 1ll * min(k - 1, 2 * n - 1 - k) * pw[k]) % mod;
  return ret * powM(pw[n]) % mod;
}
int S(int n) { return (1ll * (n - 1) * n / 2) % mod; }
int calc2() {
  int ret = 0;
  for (int t = 1; t < n; t++)
    if (mu[t])
      for (int d = 1; d * t < n; d++)
        ret = (ret + 1ll * mu[t] * pw[d] * S((d + n - 1) / (d * t))) % mod;
  return ret;
}
int o[100500];
int calc3() {
  int ret = 0;
  for (int t = 1; t < n; t++) {
    for (int i = 0; i < sd[t].size(); i++) {
      int d = sd[t][i];
      if (mu[t / d]) {
        int k = (d + n - 1) / t;
        o[2] += mu[t / d];
        o[k + 1] -= k * mu[t / d];
        o[k + 2] += (k - 1) * mu[t / d];
      }
    }
    for (int i = 2; i <= (n - 1) / t + 1; i++) o[i] = (o[i] + o[i - 1]) % mod;
    for (int i = 2; i <= (n - 1) / t + 1; i++) {
      o[i] = (o[i] + o[i - 1]) % mod;
      ret = (ret + 1ll * pw[i * t] * o[i]) % mod;
    }
    for (int i = 2; i <= (n - 1) / t + 1; i++) o[i] = 0;
  }
  return ret * powM(pw[n]) % mod;
}
int main() {
  scanf("%d%d", &n, &c);
  pw[0] = 1;
  for (int i = 1; i <= n + n; i++) pw[i] = 1ll * pw[i - 1] * c % mod;
  sieve(n);
  long long ans = calc1() + calc2() - calc3();
  ans = ans * powM(pw[n]) % mod;
  printf("%lld", (ans + mod) % mod);
  return 0;
}