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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long bin_pow(long long a, long long p) {
  p = (p % (mod - 1) + mod - 1) % (mod - 1);
  long long r = 1;
  while (p) {
    if (p & 1) r = r * a % mod;
    a = a * a % mod;
    p >>= 1;
  }
  return r;
}
long long inv(long long a) { return bin_pow(a, mod - 2); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  long long ans = 0;
  vector<long long> kpow = {bin_pow(k, -n)};
  for (int i = 1; i < n * 2; i++) kpow.push_back(kpow.back() * k % mod);
  vector<int> mu(n, 1), isprime(n, 1);
  for (long long i = 2; i < n; i++) {
    if (!isprime[i]) continue;
    for (long long j = 0; j < n; j += i) mu[j] *= -1, isprime[j] = 0;
    for (long long j = 0; j < n; j += i * i) mu[j] = 0;
  }
  for (int s = 2; s < n * 2; s++) {
    (ans += kpow[s] * min(s - 1, n * 2 - s - 1)) %= mod;
  }
  for (int g = 1; g < n; g++) {
    for (int d = 1; d <= n / (2 * g) + 1; d++) {
      for (int s = 2; s <= (n + g - 1) / (g * d); s++) {
        (ans += mu[d] * (kpow[g + n] - kpow[d * g * s]) * (s - 1)) %= mod;
      }
    }
  }
  cout << (ans * bin_pow(k, mod - 1 - n) % mod + mod) % mod << endl;
}