#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 1;
const int mod = 1e9 + 7;
int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
int mul(int a, int b) { return 1ll * a * b % mod; }
int Pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
vector<int> p[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int k;
  cin >> k;
  for (int i = 2; i < N; ++i)
    if (p[i].empty())
      for (int j = i; j < N; j += i) p[j].push_back(i);
  int ans = 0;
  for (int i = 1; i < n; ++i) ans = add(ans, Pow(k, mod - 1 + n - i));
  for (int g = 1; g < n; ++g)
    for (int s = 3; s < n + n; ++s) {
      if (s * g > n + n - 2) break;
      int l = max(1, s - (n - 1) / g) - 1;
      int r = min(s - 1, (n - 1) / g);
      if (l >= r) continue;
      int cur = 0;
      for (int m = 0; m < (1 << p[s].size()); ++m) {
        int d = 1;
        for (int i = 0; i < p[s].size(); ++i)
          if (m >> i & 1) d *= p[s][i];
        int x = r / d - l / d;
        if (__builtin_popcount(m) & 1)
          cur -= x;
        else
          cur += x;
      }
      ans = add(ans, mul(Pow(k, max(s * g - n, g)), cur));
    }
  cout << mul(ans, Pow(k, mod - 1 - n)) << endl;
}

### 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 N = 2e5 + 1;
const int mod = 1e9 + 7;
int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
int mul(int a, int b) { return 1ll * a * b % mod; }
int Pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
vector<int> p[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int k;
  cin >> k;
  for (int i = 2; i < N; ++i)
    if (p[i].empty())
      for (int j = i; j < N; j += i) p[j].push_back(i);
  int ans = 0;
  for (int i = 1; i < n; ++i) ans = add(ans, Pow(k, mod - 1 + n - i));
  for (int g = 1; g < n; ++g)
    for (int s = 3; s < n + n; ++s) {
      if (s * g > n + n - 2) break;
      int l = max(1, s - (n - 1) / g) - 1;
      int r = min(s - 1, (n - 1) / g);
      if (l >= r) continue;
      int cur = 0;
      for (int m = 0; m < (1 << p[s].size()); ++m) {
        int d = 1;
        for (int i = 0; i < p[s].size(); ++i)
          if (m >> i & 1) d *= p[s][i];
        int x = r / d - l / d;
        if (__builtin_popcount(m) & 1)
          cur -= x;
        else
          cur += x;
      }
      ans = add(ans, mul(Pow(k, max(s * g - n, g)), cur));
    }
  cout << mul(ans, Pow(k, mod - 1 - n)) << endl;
}

```