#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
inline int kpow(int a, int b) {
  int s = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) s = 1ll * s * a % mod;
  return s;
}
int n, k, su[200020], sunum, ans, po[200020], mu[200020];
bool bj[200020];
void init() {
  po[0] = 1;
  for (int i = 1; i <= 200000; ++i) po[i] = 1ll * po[i - 1] * k % mod;
  bj[1] = 1;
  mu[1] = 1;
  for (int i = 2; i <= 200000; ++i) {
    if (!bj[i]) su[++sunum] = i, mu[i] = mod - 1;
    for (int j = 1; j <= sunum && su[j] * i <= 200000; ++j) {
      bj[su[j] * i] = 1;
      if (i % su[j] == 0) break;
      mu[i * su[j]] = 1ll * mu[i] * (mod - 1) % mod;
    }
  }
}
int main() {
  scanf("%d %d", &n, &k);
  init();
  for (int d = 1; d <= n; ++d)
    if (mu[d])
      for (int sg = d; sg <= 2 * n - 2; sg += d)
        for (int s = sg; s <= 2 * n - 2; s += sg) {
          int g = s / sg, l = max(sg - (n - 1) / g, 1) - 1,
              r = min(sg - 1, (n - 1) / g);
          if (l < r) {
            if (sg <= n / g + 1)
              ans = (ans + 1ll * po[g] * (r / d - l / d) % mod * mu[d]) % mod;
            else
              ans =
                  (ans + 1ll * po[s - n] * (r / d - l / d) % mod * mu[d]) % mod;
          }
        }
  ans = 1ll * ans * kpow(po[n], mod - 2) % mod;
  printf("%d", ans);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 = 1000000007;
inline int kpow(int a, int b) {
  int s = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) s = 1ll * s * a % mod;
  return s;
}
int n, k, su[200020], sunum, ans, po[200020], mu[200020];
bool bj[200020];
void init() {
  po[0] = 1;
  for (int i = 1; i <= 200000; ++i) po[i] = 1ll * po[i - 1] * k % mod;
  bj[1] = 1;
  mu[1] = 1;
  for (int i = 2; i <= 200000; ++i) {
    if (!bj[i]) su[++sunum] = i, mu[i] = mod - 1;
    for (int j = 1; j <= sunum && su[j] * i <= 200000; ++j) {
      bj[su[j] * i] = 1;
      if (i % su[j] == 0) break;
      mu[i * su[j]] = 1ll * mu[i] * (mod - 1) % mod;
    }
  }
}
int main() {
  scanf("%d %d", &n, &k);
  init();
  for (int d = 1; d <= n; ++d)
    if (mu[d])
      for (int sg = d; sg <= 2 * n - 2; sg += d)
        for (int s = sg; s <= 2 * n - 2; s += sg) {
          int g = s / sg, l = max(sg - (n - 1) / g, 1) - 1,
              r = min(sg - 1, (n - 1) / g);
          if (l < r) {
            if (sg <= n / g + 1)
              ans = (ans + 1ll * po[g] * (r / d - l / d) % mod * mu[d]) % mod;
            else
              ans =
                  (ans + 1ll * po[s - n] * (r / d - l / d) % mod * mu[d]) % mod;
          }
        }
  ans = 1ll * ans * kpow(po[n], mod - 2) % mod;
  printf("%d", ans);
  return 0;
}

```