#include <bits/stdc++.h>
using namespace std;
const int Mod = 1e9 + 7, maxn = 1e5 + 5;
int n, k, pw[maxn], mb[maxn], ans;
bool vis[maxn];
int power(int b, int e) {
  if (!e) return 1;
  int tmp = power(b, e >> 1);
  tmp = 1ll * tmp * tmp % Mod;
  if (e & 1) tmp = 1ll * tmp * b % Mod;
  return tmp;
}
int main() {
  cin >> n >> k;
  pw[0] = 1;
  pw[1] = power(k, Mod - 2);
  for (int i = 2; i <= n; i++) pw[i] = 1ll * pw[1] * pw[i - 1] % Mod;
  for (int i = 1; i <= n; i++) mb[i] = 1;
  for (int i = 2; i <= n; i++)
    if (!vis[i]) {
      for (int j = i; j <= n; j += i) {
        vis[j] = true;
        if (j / i % i)
          mb[j] = -mb[j];
        else
          mb[j] = 0;
      }
    }
  for (int i = 1; i <= n; i++) mb[i] = (mb[i] + Mod) % Mod;
  for (int i = 1; i <= n; i++)
    for (int t = i; t <= n * 2; t += i)
      for (int g = 1; g * t <= n * 2; g++) {
        int nl = max(1, t - (n - 1) / g) - 1, nr = min((n - 1) / g, t - 1);
        if (nl < nr)
          ans = (ans + 1ll * pw[n - max(g * t - n, g)] * mb[i] % Mod *
                           (nr / i - nl / i)) %
                Mod;
      }
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int Mod = 1e9 + 7, maxn = 1e5 + 5;
int n, k, pw[maxn], mb[maxn], ans;
bool vis[maxn];
int power(int b, int e) {
  if (!e) return 1;
  int tmp = power(b, e >> 1);
  tmp = 1ll * tmp * tmp % Mod;
  if (e & 1) tmp = 1ll * tmp * b % Mod;
  return tmp;
}
int main() {
  cin >> n >> k;
  pw[0] = 1;
  pw[1] = power(k, Mod - 2);
  for (int i = 2; i <= n; i++) pw[i] = 1ll * pw[1] * pw[i - 1] % Mod;
  for (int i = 1; i <= n; i++) mb[i] = 1;
  for (int i = 2; i <= n; i++)
    if (!vis[i]) {
      for (int j = i; j <= n; j += i) {
        vis[j] = true;
        if (j / i % i)
          mb[j] = -mb[j];
        else
          mb[j] = 0;
      }
    }
  for (int i = 1; i <= n; i++) mb[i] = (mb[i] + Mod) % Mod;
  for (int i = 1; i <= n; i++)
    for (int t = i; t <= n * 2; t += i)
      for (int g = 1; g * t <= n * 2; g++) {
        int nl = max(1, t - (n - 1) / g) - 1, nr = min((n - 1) / g, t - 1);
        if (nl < nr)
          ans = (ans + 1ll * pw[n - max(g * t - n, g)] * mb[i] % Mod *
                           (nr / i - nl / i)) %
                Mod;
      }
  cout << ans << endl;
  return 0;
}

```