#include <bits/stdc++.h>
using namespace std;
const int mxN = 1e5, M = 1e9 + 7;
int n, k;
long long pk[mxN], mb[mxN], ans;
bool c[mxN];
long long iv(long long a, long long m) {
  return a < 2 ? 1 : (1 - iv(m % a, a) * m) / a + m;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  pk[1] = iv(k, M);
  for (int i = 2; i < n; ++i) pk[i] = pk[i - 1] * pk[1] % M;
  fill(mb, mb + n, 1);
  for (int i = 2; i < n; ++i) {
    if (c[i]) continue;
    for (int j = i; j < n; j += i) {
      if (j / i % i)
        mb[j] = mb[j] * (M - 1) % M;
      else
        mb[j] = 0;
      c[j] = 1;
    }
  }
  for (int i = 1; i < n; ++i) {
    if (!mb[i]) continue;
    for (int t = i; t < 2 * n - 1; t += i) {
      for (int s = t; s < 2 * n - 1; s += t) {
        int g = s / t, l = max(t - (n - 1) / g - 1, 0),
            r = min(t - 1, (n - 1) / g);
        if (l < r && g < n)
          ans = (ans + (r / i - l / i) * pk[n - max(s - n, g)] % M * mb[i]) % M;
      }
    }
  }
  cout << ans;
}

### Prompt

Generate a CPP solution to 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 int mxN = 1e5, M = 1e9 + 7;
int n, k;
long long pk[mxN], mb[mxN], ans;
bool c[mxN];
long long iv(long long a, long long m) {
  return a < 2 ? 1 : (1 - iv(m % a, a) * m) / a + m;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  pk[1] = iv(k, M);
  for (int i = 2; i < n; ++i) pk[i] = pk[i - 1] * pk[1] % M;
  fill(mb, mb + n, 1);
  for (int i = 2; i < n; ++i) {
    if (c[i]) continue;
    for (int j = i; j < n; j += i) {
      if (j / i % i)
        mb[j] = mb[j] * (M - 1) % M;
      else
        mb[j] = 0;
      c[j] = 1;
    }
  }
  for (int i = 1; i < n; ++i) {
    if (!mb[i]) continue;
    for (int t = i; t < 2 * n - 1; t += i) {
      for (int s = t; s < 2 * n - 1; s += t) {
        int g = s / t, l = max(t - (n - 1) / g - 1, 0),
            r = min(t - 1, (n - 1) / g);
        if (l < r && g < n)
          ans = (ans + (r / i - l / i) * pk[n - max(s - n, g)] % M * mb[i]) % M;
      }
    }
  }
  cout << ans;
}

```