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 ll = long long;
constexpr int MOD = 1e9 + 7;
inline int quick_pow(int a, int n) {
  int res = 1;
  while (n) {
    if (n & 1) res = (ll)res * a % MOD;
    a = (ll)a * a % MOD;
    n >>= 1;
  }
  return res;
}
inline void Add(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
  if (x < 0) x += MOD;
}
int phi[200005], mu[200005], pri[200005], tot;
int sum[200005], ans[200005];
bool np[200005];
std::vector<int> fac[200005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  phi[1] = mu[1] = 1;
  for (int i = 2; i <= n * 2; i++) {
    if (!np[i]) {
      phi[i] = i - 1;
      mu[i] = -1;
      pri[tot++] = i;
    }
    for (int j = 0; j < tot && i * pri[j] <= n * 2; j++) {
      np[i * pri[j]] = true;
      if (!(i % pri[j])) {
        phi[i * pri[j]] = phi[i] * pri[j];
        mu[i * pri[j]] = 0;
        break;
      }
      phi[i * pri[j]] = phi[i] * (pri[j] - 1);
      mu[i * pri[j]] = -mu[i];
    }
  }
  for (int i = 1; i <= n * 2; i++) {
    for (int j = 1; i * j <= n * 2; j++) fac[i * j].push_back(i);
  }
  for (int i = 1; i < n; i++) {
    for (int x : fac[i]) {
      for (int y : fac[x]) Add(ans[y], (n - i) / x * mu[x / y]);
    }
  }
  for (int i = n + 1; i < n * 2 - 1; i++) {
    for (int x : fac[i]) {
      for (int y : fac[x]) Add(sum[y], ((n - 1) / x - (i - n) / x) * mu[x / y]);
    }
    for (int x : fac[i]) {
      Add(ans[std::max(x, i - n)], sum[x]);
      sum[x] = 0;
    }
  }
  int out = 0;
  for (int i = 0, cur = 1; i <= 2 * n; i++, cur = (ll)cur * k % MOD)
    out = (out + (ll)cur * ans[i]) % MOD;
  printf("%lld\n", (ll)out * quick_pow(k, (ll)(MOD - 2) * n % (MOD - 1)) % MOD);
  return 0;
}