#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long n, c, m, phi[300000], pre[300000], pr[300000], ans, vis[300000],
    pw[300000];
void upd(long long &x, long long y) { x = (x + y) % mod; }
long long inv(long long x) {
  return x == 1 ? 1 : (mod - mod / x) * inv(mod % x) % mod;
}
int main() {
  scanf("%lld%lld", &n, &c);
  phi[1] = 1;
  for (long long i = 2; i <= n; ++i) {
    if (!vis[i]) {
      pr[++m] = i;
      phi[i] = i - 1;
    }
    for (long long j = 1; j <= m; ++j) {
      if (i * pr[j] > n) break;
      vis[i * pr[j]] = pr[j];
      if (i % pr[j] == 0) {
        phi[i * pr[j]] = phi[i] * pr[j];
        break;
      }
      phi[i * pr[j]] = phi[i] * (pr[j] - 1);
    }
    pre[i - 1] = (pre[i - 2] + phi[i]) % mod;
  }
  pw[0] = 1;
  for (long long i = 1; i <= n * 2; ++i) pw[i] = pw[i - 1] * c % mod;
  for (long long d = 1; d < n; ++d) {
    upd(ans, pw[d] * pre[n / d - 1]);
  }
  for (long long k = n + 1; k <= (n - 1) * 2; ++k) {
    upd(ans, pw[k - n] * ((n - 1) * 2 - k + 1));
  }
  for (long long d = 1; d < n; ++d) {
    long long k = (n - 1) / d + 1;
    if (n / d < k) upd(ans, (pw[d] - pw[k * d - n]) * phi[k]);
  }
  ans = ans * inv(pw[n]) % mod;
  upd(ans, mod);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

In Cpp, your task is to solve the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long n, c, m, phi[300000], pre[300000], pr[300000], ans, vis[300000],
    pw[300000];
void upd(long long &x, long long y) { x = (x + y) % mod; }
long long inv(long long x) {
  return x == 1 ? 1 : (mod - mod / x) * inv(mod % x) % mod;
}
int main() {
  scanf("%lld%lld", &n, &c);
  phi[1] = 1;
  for (long long i = 2; i <= n; ++i) {
    if (!vis[i]) {
      pr[++m] = i;
      phi[i] = i - 1;
    }
    for (long long j = 1; j <= m; ++j) {
      if (i * pr[j] > n) break;
      vis[i * pr[j]] = pr[j];
      if (i % pr[j] == 0) {
        phi[i * pr[j]] = phi[i] * pr[j];
        break;
      }
      phi[i * pr[j]] = phi[i] * (pr[j] - 1);
    }
    pre[i - 1] = (pre[i - 2] + phi[i]) % mod;
  }
  pw[0] = 1;
  for (long long i = 1; i <= n * 2; ++i) pw[i] = pw[i - 1] * c % mod;
  for (long long d = 1; d < n; ++d) {
    upd(ans, pw[d] * pre[n / d - 1]);
  }
  for (long long k = n + 1; k <= (n - 1) * 2; ++k) {
    upd(ans, pw[k - n] * ((n - 1) * 2 - k + 1));
  }
  for (long long d = 1; d < n; ++d) {
    long long k = (n - 1) / d + 1;
    if (n / d < k) upd(ans, (pw[d] - pw[k * d - n]) * phi[k]);
  }
  ans = ans * inv(pw[n]) % mod;
  upd(ans, mod);
  printf("%lld\n", ans);
  return 0;
}

```