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 = 2e5 + 10;
const int mod = 1e9 + 7;
int n, k, mu[N], pw[N];
vector<int> f[N], g[N];
int qpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % mod;
    a = 1ll * a * a % mod, b >>= 1;
  }
  return ret;
}
int C2(int n) {
  if (n < 2) return 0;
  return (1ll * n * (n - 1) / 2) % mod;
}
int main() {
  cin >> n >> k;
  mu[1] = 1;
  for (int i = 1; i < n; i++)
    for (int j = i + i; j < n; j += i) mu[j] -= mu[i];
  pw[0] = 1;
  for (int i = 1; i <= n + n; i++) pw[i] = 1ll * pw[i - 1] * k % mod;
  int inv1 = qpow(pw[n], mod - 2), inv2 = qpow(pw[n + n], mod - 2);
  for (int i = 1; i < n; i++) {
    f[i].push_back(1), g[i].push_back(0);
    for (int j = 1, t = 1; 1ll * i * j < n + n; j++)
      t = 1ll * t * pw[i] % mod, f[i].push_back((f[i].back() + t) % mod),
      g[i].push_back((g[i].back() + 1ll * j * t) % mod);
  }
  int ans = 0;
  for (int d = 1; d < n; d++)
    for (int p = 1; 1ll * p * d < n; p++) {
      int lim = (n - 1) / (p * d), cmp = (n - 1 + d) / (p * d),
          s1 = (1ll * C2(cmp) + (mod - 2ll) * C2(cmp - lim) +
                C2(cmp - 2 * lim)) %
               mod * pw[d] % mod * inv1 % mod;
      int s2 = ((1ll + 2 * lim) * (f[p * d][2 * lim] - f[p * d][cmp] + mod) +
                (mod - 1ll) * (g[p * d][2 * lim] - g[p * d][cmp] + mod)) %
               mod * inv2 % mod;
      ans = (ans + 1ll * (mod + mu[p]) * (s1 + s2)) % mod;
    }
  cout << ans;
  return 0;
}