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;
namespace factor {
vector<int> lp, pr;
void init(int n) {
  lp.resize(n + 1);
  for (int i = 2; i < n; i++) {
    if (!lp[i]) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (int p : pr) {
      if (p > lp[i] || p * i > n) break;
      lp[p * i] = p;
    }
  }
}
vector<pair<int, int>> fac(int x) {
  vector<pair<int, int>> ans;
  while (x > 1) {
    int p = lp[x];
    int m = 0;
    while (x % p == 0) {
      m++;
      x /= p;
    }
    ans.push_back({p, m});
  }
  return ans;
}
vector<int> divs(int x) {
  vector<int> ans{1};
  while (x > 1) {
    int p = lp[x];
    int m = 0;
    while (x % p == 0) {
      m++;
      x /= p;
    }
    int n = ans.size();
    ans.resize(n + n * m);
    for (int f = p, i = 1; i <= m; f *= p, i++)
      for (int j = 0; j < n; j++) ans[j + i * n] = ans[j] * f;
  }
  return ans;
}
}  // namespace factor
long long mod = 1e9 + 7;
void add(long long& a, long long b) { a = (a + b) % mod; }
void mul(long long& a, long long b) { a = (a * b) % mod; }
long long inv(long long x) {
  int p = mod - 2;
  long long ans = 1;
  while (p) {
    if (p & 1) mul(ans, x);
    mul(x, x);
    p /= 2;
  }
  return ans;
}
int main() {
  int n;
  long long k;
  cin >> n >> k;
  factor::init(2 * n);
  vector<long long> kpow(n + 1);
  kpow[0] = 1;
  for (int i = 0; i < n; i++) kpow[i + 1] = kpow[i] * k % mod;
  long long E = 0;
  for (int p = 1; p < n; p++) add(E, kpow[p]);
  for (int s = 3; s <= 2 * n - 3; s++) {
    auto divs = factor::divs(s);
    divs.pop_back();
    int ds = divs.size();
    vector<int> nvm(ds);
    int tot = 0;
    for (int i = ds; i--;) {
      int d = divs[i];
      int cnt = min(s - 1, n - 1) / d - max(s - n, 0) / d - nvm[i];
      if (s & 1 || d != s / 2) add(E, cnt * kpow[max(s - n, d)]);
      for (int j = 0; j < i; j++)
        if (d % divs[j] == 0) nvm[j] += cnt;
    }
  }
  mul(E, inv(kpow[n]));
  cout << E << '\n';
}