#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5, P = 1e9 + 7;
int n, k, ans, phi[N], s[N], v[N], p[N], pw[N];
int inv(int x) { return x == 1 ? x : 1LL * (P - P / x) * inv(P % x) % P; }
int main() {
  scanf("%d%d", &n, &k), phi[1] = s[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!v[i]) p[++p[0]] = i, phi[i] = i - 1;
    for (int j = 1; j <= p[0] && i * p[j] <= n; ++j) {
      v[i * p[j]] = 1, phi[i * p[j]] = phi[i] * (p[j] - 1);
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      }
    }
    s[i] = (s[i - 1] + phi[i]) % P;
  }
  for (int i = pw[0] = 1; i <= n; ++i) pw[i] = 1LL * pw[i - 1] * k % P;
  pw[0] = 0;
  for (int i = 1; i < n - 1; ++i) (ans += 1LL * pw[i] * (n - i - 1) % P) %= P;
  for (int i = 1; i < n; ++i)
    (ans += 1LL * (s[(n - 1) / i + 1] - s[1] + P) * pw[i] % P) %= P;
  for (int i = 1, x; i < n; ++i)
    x = (n - 1) / i + 1, (ans += P - 1LL * phi[x] * pw[x * i - n] % P) %= P;
  printf("%lld", 1LL * ans * inv(pw[n]) % P);
  return 0;
}

### 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 N = 1e5 + 5, P = 1e9 + 7;
int n, k, ans, phi[N], s[N], v[N], p[N], pw[N];
int inv(int x) { return x == 1 ? x : 1LL * (P - P / x) * inv(P % x) % P; }
int main() {
  scanf("%d%d", &n, &k), phi[1] = s[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!v[i]) p[++p[0]] = i, phi[i] = i - 1;
    for (int j = 1; j <= p[0] && i * p[j] <= n; ++j) {
      v[i * p[j]] = 1, phi[i * p[j]] = phi[i] * (p[j] - 1);
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      }
    }
    s[i] = (s[i - 1] + phi[i]) % P;
  }
  for (int i = pw[0] = 1; i <= n; ++i) pw[i] = 1LL * pw[i - 1] * k % P;
  pw[0] = 0;
  for (int i = 1; i < n - 1; ++i) (ans += 1LL * pw[i] * (n - i - 1) % P) %= P;
  for (int i = 1; i < n; ++i)
    (ans += 1LL * (s[(n - 1) / i + 1] - s[1] + P) * pw[i] % P) %= P;
  for (int i = 1, x; i < n; ++i)
    x = (n - 1) / i + 1, (ans += P - 1LL * phi[x] * pw[x * i - n] % P) %= P;
  printf("%lld", 1LL * ans * inv(pw[n]) % P);
  return 0;
}

```