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>
const int MOD = 1000000007;
const int MAX_N = 100000;
void selfMult(int &a, int b) { a = (long long)a * b % MOD; }
int mult(int a, int b) { return (long long)a * b % MOD; }
void selfAdd(int &a, int b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
}
int power(int base, int exp) {
  int answer = 1;
  while (exp > 0) {
    if (exp % 2 == 1) {
      selfMult(answer, base);
    }
    selfMult(base, base);
    exp /= 2;
  }
  return answer;
}
int inv(int value) { return power(value, MOD - 2); }
int gcd(int a, int b) {
  int r = a % b;
  while (r > 0) {
    a = b;
    b = r;
    r = a % b;
  }
  return b;
}
int di[1 + 2 * MAX_N];
int coprime(int n, int min, int max) {
  std::vector<int> primes;
  primes.push_back(1);
  while (n > 1) {
    int p = di[n];
    while (di[n] == p) {
      n /= p;
    }
    int size = primes.size();
    for (int i = 0; i < size; i++) {
      primes.push_back(primes[i] * -p);
    }
  }
  min--;
  int answer = 0;
  for (int p : primes) {
    answer += max / p - min / p;
  }
  return answer;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int freedoms[1 + MAX_N];
  freedoms[0] = 1;
  freedoms[1] = k;
  for (int i = 2; i <= n; i++) {
    freedoms[i] = freedoms[i - 1];
    selfMult(freedoms[i], k);
  }
  for (int i = 0; i <= 2 * n; i++) {
    di[i] = 0;
  }
  di[1] = 1;
  for (int p = 2; p <= 2 * n; p++) {
    if (di[p] == 0) {
      for (int m = p; m <= 2 * n; m += p) {
        if (di[m] == 0) {
          di[m] = p;
        }
      }
    }
  }
  int answer = 0;
  for (int s = 3; s <= 2 * n - 3; s++) {
    int no = s;
    std::vector<int> divs;
    divs.clear();
    divs.push_back(1);
    while (no > 1) {
      int exp = 0;
      int p = di[no];
      while (di[no] == p) {
        no /= p;
        exp++;
      }
      int po = 1;
      int size = divs.size();
      for (int e = 0; e < exp; e++) {
        po *= p;
        for (int i = 0; i < size; i++) {
          divs.push_back(divs[i] * po);
        }
      }
    }
    for (int gcd : divs) {
      if (gcd < s) {
        int min = std::max(1, (s - n + 1 + gcd - 1) / gcd);
        int max = std::min(n - 1, (s - 1) / 2) / gcd;
        selfAdd(answer, mult(coprime(s / gcd, min, max),
                             freedoms[std::max(s - n, gcd)]));
      }
    }
  }
  selfMult(answer, 2);
  for (int i = 1; i < n; i++) {
    selfAdd(answer, freedoms[n - i]);
  }
  selfMult(answer, power(inv(k), n));
  printf("%d\n", answer);
  return 0;
}