#include <bits/stdc++.h>
using namespace std;
namespace FGF {
int n, m;
const int N = 2e5 + 5, mo = 1e9 + 7;
int p[N], phi[N], s[N], cnt, ans, c[N];
int qpow(int x, int y) {
  int s = 1;
  while (y) {
    if (y & 1) s = 1ll * s * x % mo;
    x = 1ll * x * x % mo;
    y >>= 1;
  }
  return s;
}
void init() {
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!phi[i]) p[++cnt] = i, phi[i] = i - 1;
    for (int j = 1; j <= cnt && p[j] * i <= n; j++) {
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      }
      phi[i * p[j]] = phi[i] * (p[j] - 1);
    }
  }
}
void work() {
  scanf("%d%d", &n, &m);
  init();
  for (int i = 2; i <= n; i++) s[i] = (s[i - 1] + phi[i]) % mo;
  for (int i = 1; i < n - 1; i++) c[i] = n - i - 1;
  for (int i = 1, d; i < n; i++)
    d = (n - 1) / i, c[i] = (c[i] + s[d + 1]) % mo,
    c[(d + 1) * i - n] = (c[(d + 1) * i - n] - phi[d + 1] + mo) % mo;
  for (int i = 1, k = m; i < n; i++, k = 1ll * k * m % mo)
    ans = (ans + 1ll * c[i] * k % mo) % mo;
  printf("%lld", 1ll * ans * qpow(m, mo - 1 - n) % mo);
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}

### Prompt

Create a solution in CPP for 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;
namespace FGF {
int n, m;
const int N = 2e5 + 5, mo = 1e9 + 7;
int p[N], phi[N], s[N], cnt, ans, c[N];
int qpow(int x, int y) {
  int s = 1;
  while (y) {
    if (y & 1) s = 1ll * s * x % mo;
    x = 1ll * x * x % mo;
    y >>= 1;
  }
  return s;
}
void init() {
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!phi[i]) p[++cnt] = i, phi[i] = i - 1;
    for (int j = 1; j <= cnt && p[j] * i <= n; j++) {
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      }
      phi[i * p[j]] = phi[i] * (p[j] - 1);
    }
  }
}
void work() {
  scanf("%d%d", &n, &m);
  init();
  for (int i = 2; i <= n; i++) s[i] = (s[i - 1] + phi[i]) % mo;
  for (int i = 1; i < n - 1; i++) c[i] = n - i - 1;
  for (int i = 1, d; i < n; i++)
    d = (n - 1) / i, c[i] = (c[i] + s[d + 1]) % mo,
    c[(d + 1) * i - n] = (c[(d + 1) * i - n] - phi[d + 1] + mo) % mo;
  for (int i = 1, k = m; i < n; i++, k = 1ll * k * m % mo)
    ans = (ans + 1ll * c[i] * k % mo) % mo;
  printf("%lld", 1ll * ans * qpow(m, mo - 1 - n) % mo);
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}

```