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 Maxn = 10000005, p = 1e9 + 7;
int n, k, cnt, phi[Maxn], prim[Maxn], sum[Maxn], ct[Maxn];
long long ans;
bool vis[Maxn];
void init(void) {
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) prim[++cnt] = i, phi[i] = i - 1;
    for (int j = 1; j <= cnt && i * prim[j] <= n; j++) {
      vis[i * prim[j]] = true;
      if (i % prim[j] == 0) {
        phi[i * prim[j]] = phi[i] * prim[j];
        break;
      }
      phi[i * prim[j]] = phi[i] * (prim[j] - 1);
    }
  }
}
long long get_inv(int x) {
  if (x <= 1) return 1;
  return (p - p / x) * get_inv(p % x) % p;
}
int main() {
  scanf("%d%d", &n, &k);
  init();
  for (int i = 2; i <= n; i++) sum[i] = (sum[i - 1] + phi[i]) % p;
  for (int i = 1; i <= n; i++) ct[i] = n - i - 1;
  for (int i = 1; i < n; i++) {
    int l = (n - 1) / i;
    (ct[i] += sum[l + 1]) %= p;
    (ct[(l + 1) * i - n] += p - phi[l + 1]) %= p;
  }
  long long now = 1;
  for (int i = 1; i < n; i++) {
    now = now * k % p;
    (ans += ct[i] * now) %= p;
  }
  printf("%lld", ans * get_inv(now * k % p) % p);
  return 0;
}