#include <bits/stdc++.h>
using namespace std;
namespace FGF {
int n, m;
const int N = 2e5 + 5, mo = 1e9 + 7;
int vis[N], mu[N], p[N], cnt, pw[N], ans;
vector<int> g[N];
int inv(int x) {
  int s = 1, y = mo - 2;
  while (y) {
    if (y & 1) s = 1ll * s * x % mo;
    x = 1ll * x * x % mo;
    y >>= 1;
  }
  return s;
}
void init() {
  mu[1] = 1, pw[0] = 1, pw[1] = m;
  for (int i = 2; i <= 2 * n; i++) {
    pw[i] = 1ll * pw[i - 1] * m % mo;
    if (!vis[i]) p[++cnt] = i, mu[i] = -1;
    for (int j = 1; j <= cnt && p[j] * i <= 2 * n; j++) {
      vis[p[j] * i] = 1;
      if (i % p[j] == 0) break;
      mu[p[j] * i] = -mu[i];
    }
  }
  for (int i = 1; i <= 2 * n; i++)
    for (int j = i; j <= 2 * n; j += i) g[j].push_back(i);
}
int calc(int x, int y) {
  if (y <= 0 || x <= 1) return 0;
  int l = max(1, x - y), r = min(y, x - 1), sum = 0;
  if (l > r) return 0;
  for (int i = 0; i < g[x].size(); i++)
    sum = (sum + mu[g[x][i]] * (r / g[x][i] - (l - 1) / g[x][i]) % mo) % mo;
  return (sum + mo) % mo;
}
void work() {
  scanf("%d%d", &n, &m);
  init();
  for (int s = 2; s <= 2 * n - 2; s++)
    for (int i = 0; i < g[s].size(); i++)
      ans = (ans + 1ll * calc(s / g[s][i], (n - 1) / g[s][i]) *
                       pw[max(s - n, g[s][i])] % mo) %
            mo;
  printf("%lld", 1ll * ans * inv(pw[n]) % mo);
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}

### Prompt

Your challenge is to write 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;
namespace FGF {
int n, m;
const int N = 2e5 + 5, mo = 1e9 + 7;
int vis[N], mu[N], p[N], cnt, pw[N], ans;
vector<int> g[N];
int inv(int x) {
  int s = 1, y = mo - 2;
  while (y) {
    if (y & 1) s = 1ll * s * x % mo;
    x = 1ll * x * x % mo;
    y >>= 1;
  }
  return s;
}
void init() {
  mu[1] = 1, pw[0] = 1, pw[1] = m;
  for (int i = 2; i <= 2 * n; i++) {
    pw[i] = 1ll * pw[i - 1] * m % mo;
    if (!vis[i]) p[++cnt] = i, mu[i] = -1;
    for (int j = 1; j <= cnt && p[j] * i <= 2 * n; j++) {
      vis[p[j] * i] = 1;
      if (i % p[j] == 0) break;
      mu[p[j] * i] = -mu[i];
    }
  }
  for (int i = 1; i <= 2 * n; i++)
    for (int j = i; j <= 2 * n; j += i) g[j].push_back(i);
}
int calc(int x, int y) {
  if (y <= 0 || x <= 1) return 0;
  int l = max(1, x - y), r = min(y, x - 1), sum = 0;
  if (l > r) return 0;
  for (int i = 0; i < g[x].size(); i++)
    sum = (sum + mu[g[x][i]] * (r / g[x][i] - (l - 1) / g[x][i]) % mo) % mo;
  return (sum + mo) % mo;
}
void work() {
  scanf("%d%d", &n, &m);
  init();
  for (int s = 2; s <= 2 * n - 2; s++)
    for (int i = 0; i < g[s].size(); i++)
      ans = (ans + 1ll * calc(s / g[s][i], (n - 1) / g[s][i]) *
                       pw[max(s - n, g[s][i])] % mo) %
            mo;
  printf("%lld", 1ll * ans * inv(pw[n]) % mo);
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}

```