#include <bits/stdc++.h>
using namespace std;
const int mo = 1000000007;
const int N = 200005;
int power(int x, int y) {
  int s = 1;
  for (; y; y /= 2, x = 1ll * x * x % mo)
    if (y & 1) s = 1ll * s * x % mo;
  return s;
}
int n, k, mu[N];
vector<int> di[N];
int solve1() {
  int ans = 0;
  for (int i = (int)(1); i <= (int)(n - 1); i++)
    ans = (ans + power(k, mo - 1 - i)) % mo;
  return ans;
}
int solve2() {
  int ans = 0;
  mu[1] = 1;
  for (int i = (int)(1); i <= (int)(N - 1); i++)
    for (int j = (int)(2); j <= (int)((N - 1) / i); j++) mu[i * j] -= mu[i];
  for (int i = (int)(1); i <= (int)(N - 1); i++)
    for (int j = (int)(1); j <= (int)((N - 1) / i); j++) di[i * j].push_back(i);
  for (int i = (int)(2); i <= (int)(2 * n - 2); i++) {
    int l = max(1, i - (n - 1));
    int r = min(n - 1, i - 1);
    for (int j = (int)(0); j <= (int)(di[i].size() - 1); j++) {
      if (2 * di[i][j] == i) continue;
      int nl = (l - 1) / di[i][j] + 1;
      int nr = r / di[i][j];
      if (nl > nr) continue;
      int val = di[i][j], cnt = 0;
      int res = i / di[i][j];
      for (int k = (int)(0); k <= (int)(di[res].size() - 1); k++) {
        int v = di[res][k];
        if (!mu[v]) continue;
        int nnl = (nl - 1) / v + 1;
        int nnr = nr / v;
        if (nnl <= nnr) cnt += mu[v] * (nnr - nnl + 1);
      }
      int tms = max(i - n, val);
      ans = (ans + 1ll * power(k, tms) * cnt) % mo;
    }
  }
  ans = 1ll * ans * power(k, mo - 1 - n) % mo;
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  int ans = (solve1() + solve2()) % mo;
  printf("%d\n", ans);
}

### Prompt

Please create a solution in CPP 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 mo = 1000000007;
const int N = 200005;
int power(int x, int y) {
  int s = 1;
  for (; y; y /= 2, x = 1ll * x * x % mo)
    if (y & 1) s = 1ll * s * x % mo;
  return s;
}
int n, k, mu[N];
vector<int> di[N];
int solve1() {
  int ans = 0;
  for (int i = (int)(1); i <= (int)(n - 1); i++)
    ans = (ans + power(k, mo - 1 - i)) % mo;
  return ans;
}
int solve2() {
  int ans = 0;
  mu[1] = 1;
  for (int i = (int)(1); i <= (int)(N - 1); i++)
    for (int j = (int)(2); j <= (int)((N - 1) / i); j++) mu[i * j] -= mu[i];
  for (int i = (int)(1); i <= (int)(N - 1); i++)
    for (int j = (int)(1); j <= (int)((N - 1) / i); j++) di[i * j].push_back(i);
  for (int i = (int)(2); i <= (int)(2 * n - 2); i++) {
    int l = max(1, i - (n - 1));
    int r = min(n - 1, i - 1);
    for (int j = (int)(0); j <= (int)(di[i].size() - 1); j++) {
      if (2 * di[i][j] == i) continue;
      int nl = (l - 1) / di[i][j] + 1;
      int nr = r / di[i][j];
      if (nl > nr) continue;
      int val = di[i][j], cnt = 0;
      int res = i / di[i][j];
      for (int k = (int)(0); k <= (int)(di[res].size() - 1); k++) {
        int v = di[res][k];
        if (!mu[v]) continue;
        int nnl = (nl - 1) / v + 1;
        int nnr = nr / v;
        if (nnl <= nnr) cnt += mu[v] * (nnr - nnl + 1);
      }
      int tms = max(i - n, val);
      ans = (ans + 1ll * power(k, tms) * cnt) % mo;
    }
  }
  ans = 1ll * ans * power(k, mo - 1 - n) % mo;
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  int ans = (solve1() + solve2()) % mo;
  printf("%d\n", ans);
}

```