#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int expmod(int a, int b) {
  while (b < 0) b += mod - 1;
  int rlt = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) rlt = 1ll * rlt * a % mod;
  return rlt;
}
int runs, p[200010], phi[200010], mu[200010];
vector<int> d[200010];
void ppm() {
  phi[1] = 1, mu[1] = 1;
  d[1] = {1};
  for (int i = 2; i < 200010; i++) {
    if (!phi[i]) p[++runs] = i, phi[i] = i - 1, mu[i] = -1, d[i] = {1, i};
    for (int j = 1; j <= runs && i * p[j] < 200010; j++) {
      int k = i * p[j];
      phi[k] = phi[i] * (p[j] - 1);
      if (i % p[j] == 0) {
        d[k] = d[i];
        phi[k] = phi[i] * p[j];
        break;
      }
      int sz = d[i].size();
      for (int l = 0; l < sz; l++)
        d[k].push_back(d[i][l]), d[k].push_back(d[i][l] * p[j]);
      mu[k] = -mu[i];
    }
  }
}
int calc(int x, int up) {
  int ans = 0;
  for (int a : d[x]) ans += (up / a) * mu[a];
  return ans;
}
int n, k;
int main() {
  ppm();
  scanf("%d%d", &n, &k);
  int ans = 0;
  for (int gc = 1; gc < n; gc++) {
    for (int sum = 2 * gc; sum <= 2 * (n - 1); sum += gc) {
      int cnt = calc(sum / gc, min(sum / gc - 1, (n - 1) / gc)) -
                calc(sum / gc, max(1, sum / gc - (n - 1) / gc) - 1);
      cnt /= 2;
      ans = (ans + 1ll * cnt * expmod(k, max(gc - n, sum - n * 2))) % mod;
    }
  }
  ans *= 2;
  ans %= mod;
  for (int i = 1; i < n; i++) ans = (ans + expmod(k, -i)) % mod;
  cout << ans;
}

### Prompt

Please provide a Cpp coded solution 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 = 1e9 + 7;
int expmod(int a, int b) {
  while (b < 0) b += mod - 1;
  int rlt = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) rlt = 1ll * rlt * a % mod;
  return rlt;
}
int runs, p[200010], phi[200010], mu[200010];
vector<int> d[200010];
void ppm() {
  phi[1] = 1, mu[1] = 1;
  d[1] = {1};
  for (int i = 2; i < 200010; i++) {
    if (!phi[i]) p[++runs] = i, phi[i] = i - 1, mu[i] = -1, d[i] = {1, i};
    for (int j = 1; j <= runs && i * p[j] < 200010; j++) {
      int k = i * p[j];
      phi[k] = phi[i] * (p[j] - 1);
      if (i % p[j] == 0) {
        d[k] = d[i];
        phi[k] = phi[i] * p[j];
        break;
      }
      int sz = d[i].size();
      for (int l = 0; l < sz; l++)
        d[k].push_back(d[i][l]), d[k].push_back(d[i][l] * p[j]);
      mu[k] = -mu[i];
    }
  }
}
int calc(int x, int up) {
  int ans = 0;
  for (int a : d[x]) ans += (up / a) * mu[a];
  return ans;
}
int n, k;
int main() {
  ppm();
  scanf("%d%d", &n, &k);
  int ans = 0;
  for (int gc = 1; gc < n; gc++) {
    for (int sum = 2 * gc; sum <= 2 * (n - 1); sum += gc) {
      int cnt = calc(sum / gc, min(sum / gc - 1, (n - 1) / gc)) -
                calc(sum / gc, max(1, sum / gc - (n - 1) / gc) - 1);
      cnt /= 2;
      ans = (ans + 1ll * cnt * expmod(k, max(gc - n, sum - n * 2))) % mod;
    }
  }
  ans *= 2;
  ans %= mod;
  for (int i = 1; i < n; i++) ans = (ans + expmod(k, -i)) % mod;
  cout << ans;
}

```