#include <bits/stdc++.h>
const int mod = 1e9 + 7;
using std::cin;
using std::cout;
const int maxn = 200100;
int n, k;
inline long long pow(int a, int b, int ans = 1) {
  for (; b; b >>= 1, a = (long long)a * a % mod)
    if (b & 1) {
      ans = (long long)ans * a % mod;
    }
  return ans;
}
int val[maxn];
int cnt[maxn];
int d[maxn];
int main() {
  std::ios::sync_with_stdio(false), std::cin.tie(0);
  cin >> n >> k;
  --n;
  long long ans = 0;
  for (int gcd = 1; gcd <= n; ++gcd) {
    const int N = n / gcd;
    for (int i = 1; i <= N; ++i) d[i] = 0;
    d[1] = 1;
    for (int i = 1; i <= N; ++i) {
      for (int j = i + i; j <= N; j += i) d[j] -= d[i];
      const int M = N / i;
      for (int j = 2; j <= M + M; ++j) {
        cnt[i * j] += (j <= M ? j - 1 : M + M - j + 1) * d[i];
      }
    }
    for (int i = 2; i <= N * 2; ++i) {
      ans = (ans + cnt[i] * pow(k, std::max(gcd, i * gcd - n - 1))) % mod;
      cnt[i] = 0;
    }
  }
  cout << ans * pow(k, mod - 2 - n) % mod << '\n';
}

### 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>
const int mod = 1e9 + 7;
using std::cin;
using std::cout;
const int maxn = 200100;
int n, k;
inline long long pow(int a, int b, int ans = 1) {
  for (; b; b >>= 1, a = (long long)a * a % mod)
    if (b & 1) {
      ans = (long long)ans * a % mod;
    }
  return ans;
}
int val[maxn];
int cnt[maxn];
int d[maxn];
int main() {
  std::ios::sync_with_stdio(false), std::cin.tie(0);
  cin >> n >> k;
  --n;
  long long ans = 0;
  for (int gcd = 1; gcd <= n; ++gcd) {
    const int N = n / gcd;
    for (int i = 1; i <= N; ++i) d[i] = 0;
    d[1] = 1;
    for (int i = 1; i <= N; ++i) {
      for (int j = i + i; j <= N; j += i) d[j] -= d[i];
      const int M = N / i;
      for (int j = 2; j <= M + M; ++j) {
        cnt[i * j] += (j <= M ? j - 1 : M + M - j + 1) * d[i];
      }
    }
    for (int i = 2; i <= N * 2; ++i) {
      ans = (ans + cnt[i] * pow(k, std::max(gcd, i * gcd - n - 1))) % mod;
      cnt[i] = 0;
    }
  }
  cout << ans * pow(k, mod - 2 - n) % mod << '\n';
}

```