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;
class Solver {
 private:
  static constexpr int MAX_N = 1e5;
  static constexpr int MOD = 1e9 + 7;
  int n, k;
  int pow_k[MAX_N];
  vector<int> prime_divisors[MAX_N * 2];
  int dfs(int x, int y, int z, int l, int r, int coef) {
    if (y == prime_divisors[x].size()) {
      return coef * (r / z - (l - 1) / z);
    }
    return dfs(x, y + 1, z, l, r, coef) +
           dfs(x, y + 1, z * prime_divisors[x][y], l, r, -coef);
  }
  int calc(int x, int l, int r) { return dfs(x, 0, 1, l, r, 1); }
  int Pow(int a, int b) {
    int ans = 1;
    while (b > 0) {
      if (b & 1) {
        ans = 1ll * ans * a % MOD;
      }
      a = 1ll * a * a % MOD;
      b >>= 1;
    }
    return ans;
  }

 public:
  void solve() {
    cin >> n >> k;
    pow_k[0] = 1;
    for (int i = 1; i <= n - 1; ++i) {
      pow_k[i] = 1ll * pow_k[i - 1] * k % MOD;
    }
    int sum = 0;
    for (int i = 2; i <= n * 2 - 2; ++i) {
      int x = i;
      for (int j = 2; j * j <= x; ++j) {
        if (x % j == 0) {
          prime_divisors[i].push_back(j);
          while (x % j == 0) {
            x /= j;
          }
        }
      }
      if (x > 1) {
        prime_divisors[i].push_back(x);
      }
    }
    int ans = 0;
    for (int gcd = 1; gcd <= n - 1; ++gcd) {
      int lim = (n - 1) / gcd;
      for (int s = 2; s <= lim * 2; ++s) {
        int l = max(1, s - lim), r = min(s - 1, lim);
        int cnt = calc(s, l, r);
        ans = (ans + 1ll * cnt * pow_k[max(gcd, gcd * s - n)]) % MOD;
      }
    }
    ans = 1ll * ans * Pow(k, MOD - 1 - n) % MOD;
    cout << ans << endl;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  Solver().solve();
  return 0;
}