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 long long md = 1e9 + 7;
long long qp(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % md;
    a = a * a % md;
    b >>= 1;
  }
  return ret;
}
bool mk[200200];
int pri[200200], mu[200200];
void mobi() {
  mu[1] = 1;
  for (int i = 2; i < 200200; i++) {
    if (!mk[i]) {
      mk[i] = 1;
      pri[++pri[0]] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= pri[0] && i * pri[j] < 200200; j++) {
      mk[i * pri[j]] = 1;
      if (i % pri[j]) {
        mu[i * pri[j]] = -mu[i];
      } else {
        break;
      }
    }
  }
  return;
}
long long n, k;
long long pk[200200];
long long cnt[200200];
int main() {
  mobi();
  scanf("%lld%lld", &n, &k);
  pk[0] = 1;
  for (int i = 1; i <= n; i++) pk[i] = pk[i - 1] * k % md;
  long long ans = 0;
  for (long long g = 1; g < n; g++) {
    long long m = (n - 1) * 2 / g;
    for (long long d = 1; d <= m; d++) {
      for (long long s = d; s <= m; s += d) {
        cnt[s] += 1ll * mu[d] *
                  (min((n - 1) / g, 1ll * s - 1) / d -
                   max(0 * 1ll, 1ll * (s * g - n) / g) / d);
        cnt[s] %= md;
      }
    }
    for (long long s = 1; s <= m; s++)
      ans = (ans + 1ll * pk[max(1ll * g, 1ll * s * g - n)] * cnt[s] % md) % md;
    for (int s = 0; s <= m + 2; s++) cnt[s] = 0;
  }
  ans = ans * qp(pk[n], md - 2) % md;
  printf("%lld\n", ans);
  return 0;
}