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 MOD = 1e9 + 7;
long long qp(long long a, long long b) {
  long long x = 1;
  a %= MOD;
  b %= (MOD - 1);
  while (b < 0) b += MOD - 1;
  while (b) {
    if (b & 1) x = x * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return x;
}
int mu[666666];
bool ip[666666];
long long ps(long long k, long long n) {
  if (k == 1) return n + 1;
  return (qp(k, n + 1) - 1) * qp(k - 1, MOD - 2) % MOD;
}
long long ps(long long k, long long l, long long r) {
  return qp(k, l) * ps(k, r - l) % MOD;
}
long long pp(long long k, long long n) {
  return (n * qp(k, n + 1) - ps(k, 1, n)) % MOD * qp(k - 1, MOD - 2) % MOD;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) ans += i * qp(k, i), ans %= MOD;
  return ans;
}
long long su(long long l, long long r) {
  l %= MOD;
  r %= MOD;
  return (l + r) * (r - l + 1) / 2 % MOD;
}
int main() {
  int n, k;
  cin >> n >> k;
  if (n == 1) {
    puts("0");
    return 0;
  }
  if (k == 1) {
    long long w = n - 1;
    w = w * w % MOD;
    cout << w << "\n";
    return 0;
  }
  for (int i = 1; i <= n; ++i) mu[i] = 1;
  for (int i = 2; i <= n; ++i)
    if (!ip[i]) {
      for (int j = i; j <= n; j += i) {
        ip[j] = 1;
        if ((j / i) % i == 0)
          mu[j] = 0;
        else
          mu[j] = -mu[j];
      }
    }
  long long ans = 0;
  for (int s = 1; s <= n; ++s) {
    int u = (n + s) / s, r = (n - 1) / s;
    long long cd = 0, ap = 0;
    for (int x = 1; x <= r; ++x) {
      long long ux = u / x, rx = r / x;
      long long K = qp(k, x * s);
      long long R = min(ux, rx + 1);
      cd += mu[x] * su(1, R - 1);
      ap += mu[x] * K * pp(K, R - 1);
      cd += mu[x] * rx * (ux - R);
      ap += mu[x] * rx * ps(K, R + 1, ux);
      cd %= MOD;
      ap %= MOD;
    }
    cd %= MOD, ap %= MOD;
    ans += qp(k, s) * cd;
    ans -= ap * qp(k, -n);
    ans %= MOD;
  }
  for (int i = 1; i < n; ++i) ans += ps(k, i + 1 - n, i + n - 1 - n);
  ans = ans % MOD * qp(k, n * (long long)(MOD - 2)) % MOD;
  ans = (ans % MOD + MOD) % MOD;
  cout << ans << "\n";
}