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 vu[200010], pr[200010], vi[200010], ct, n, k, m[200010], c[200010];
void init() {
  vu[1] = 1;
  for (long long i = 2; i < 200010; i++) {
    if (!vi[i]) {
      pr[++ct] = i;
      vu[i] = -1;
    }
    for (long long j = 1; j <= ct && i * pr[j] < 200010; j++) {
      vi[i * pr[j]] = 1;
      if (i % pr[j] == 0) break;
      vu[i * pr[j]] = -vu[i];
    }
  }
}
long long qp(long long x, long long y) {
  long long r = 1;
  for (; y; y >>= 1, x = x * x % 1000000007ll)
    if (y & 1) r = r * x % 1000000007ll;
  return r;
}
signed main() {
  cin >> n >> k;
  init();
  m[0] = 1;
  long long r = 0;
  for (long long i = 1; i <= n; i++) m[i] = m[i - 1] * k % 1000000007ll;
  for (long long d = 1; d < n; d++) {
    long long l = (2 * n - 2) / d;
    for (long long t = 1; t <= l; t++)
      for (long long s = t; s <= l; s += t)
        c[s] = (c[s] + vu[t] * (min(s - 1, (n - 1) / d) / t -
                                max(0ll, (s * d - n) / d) / t)) %
               1000000007ll;
    for (long long s = 1; s <= l; s++)
      r = (r + m[max(s * d - n, d)] * c[s] % 1000000007ll) % 1000000007ll;
    for (long long i = 0; i < l + 5; i++) c[i] = 0;
  }
  r = r * (qp(qp(k, n), 1000000007ll - 2)) % 1000000007ll;
  cout << r;
}