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 = 1e5 + 10, mod = 1e9 + 7;
int u[N], k;
int vis[N], pr[N], cnt, n;
long long pw(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
long long inv[N];
int main() {
  scanf("%d%d", &n, &k);
  inv[0] = 1;
  inv[1] = pw(k, mod - 2);
  for (int i = 2; i <= n; i++) {
    inv[i] = inv[i - 1] * inv[1] % mod;
  }
  u[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) pr[++cnt] = i, u[i] = -1;
    for (int j = 1; j <= cnt && 1LL * pr[j] * i <= n; j++) {
      vis[i * pr[j]] = 1;
      if (i % pr[j]) {
        u[i * pr[j]] = -u[i];
      } else {
        u[i * pr[j]] = 0;
        break;
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n - 1; i++) {
    for (int s2 = i; s2 <= 2 * n - 1; s2 += i) {
      for (int s = s2; s <= 2 * n - 1; s += s2) {
        long long gcd = s / s2;
        long long l = max(1LL, s2 - (n - 1) / gcd),
                  r = min(s2 - 1LL, (n - 1) / gcd);
        if (l <= r)
          ans += (r / i - (l - 1) / i) * inv[n - max(s - (long long)n, gcd)] %
                 mod * (u[i] + mod) % mod;
        ans %= mod;
      }
    }
  }
  printf("%lld", ans);
  return 0;
}