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;
const int N = 100005, M = 1000000007;
int n, p, i, j, k, mu[N], vis[N];
long long mi[N], ans;
long long qpow(long long a, long long b) {
  long long s = a, ans = 1;
  while (b) {
    if (b & 1) ans = ans * s % M;
    s = s * s % M;
    b >>= 1;
  }
  return ans;
}
int main() {
  scanf("%d %d", &n, &p);
  mi[0] = 1;
  for (i = 1; i <= n; ++i) {
    mi[i] = mi[i - 1] * p % M;
    mu[i] = 1;
  }
  for (i = 2; i <= n; ++i)
    if (!vis[i])
      for (j = i; j <= n; j += i) {
        vis[j] = 1;
        mu[j] = -mu[j];
        if (j / i % i == 0) mu[j] = 0;
      }
  for (i = 1; i <= n - 1; ++i)
    for (j = 1; j <= (n - 1) / i; ++j) {
      int t = (n - 1) / i / j;
      long long s = 0;
      for (k = 2; k <= t * 2; ++k)
        s += (t - abs(t + 1 - k)) * mi[max(i * j * k - n, i)];
      ans = (ans + s * mu[j]) % M;
    }
  cout << (ans * qpow(mi[n], M - 2) % M + M) % M;
}