#include <bits/stdc++.h>
const int MOD = 1e9 + 7;
int power(int a, int n) {
  int result = 1;
  while (n) {
    if (n & 1) {
      result = 1LL * result * a % MOD;
    }
    a = 1LL * a * a % MOD;
    n >>= 1;
  }
  return result;
}
void add(int &x, int a) {
  x += a;
  if (x >= MOD) {
    x -= MOD;
  }
}
const int N = 100000;
int mu[N], mpowers[N], minvs[N];
std::vector<int> divs[N], sfdivs[N];
int gsum(int g, int n) { return (1LL + MOD - mpowers[g * n]) * minvs[g] % MOD; }
int main() {
  mu[1] = 1;
  for (int i = 1; i < N; ++i) {
    for (int j = i; j < N; j += i) {
      divs[j].push_back(i);
      if (mu[i]) {
        sfdivs[j].push_back(i);
      }
    }
    for (int j = i + i; j < N; j += i) {
      mu[j] -= mu[i];
    }
  }
  int n, m;
  scanf("%d%d", &n, &m);
  if (m == 1) {
    printf("%d\n", (int)(1LL * (n - 1) * (n - 1) % MOD));
  } else {
    mpowers[0] = 1;
    for (int i = 1; i < n; ++i) {
      mpowers[i] = 1LL * mpowers[i - 1] * m % MOD;
      minvs[i] = 1LL * mpowers[i] * power(1 + MOD - mpowers[i], MOD - 2) % MOD;
    }
    int inv_kn = power(m, 1LL * n * (MOD - 2) % (MOD - 1));
    int result = 1LL * power(gsum(1, n - 1), 2) * inv_kn % MOD;
    for (int p = 1; p < n; ++p) {
      for (int g : divs[p]) {
        int le_count = 0, le_sum = 0;
        for (int _ : sfdivs[p / g]) {
          int gg = g * _;
          int count = std::min(n - 1, n + g - p) / gg;
          le_count += mu[_] * count;
          int sum = gsum(gg, count);
          add(le_sum, mu[_] > 0 ? sum : MOD - sum);
        }
        add(result, 1LL * le_count * mpowers[g] % MOD);
        add(result, MOD - 1LL * le_sum * mpowers[p] % MOD * inv_kn % MOD);
      }
    }
    printf("%d\n", (int)(1LL * result * inv_kn % MOD));
  }
}

### Prompt

Generate a CPP solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
const int MOD = 1e9 + 7;
int power(int a, int n) {
  int result = 1;
  while (n) {
    if (n & 1) {
      result = 1LL * result * a % MOD;
    }
    a = 1LL * a * a % MOD;
    n >>= 1;
  }
  return result;
}
void add(int &x, int a) {
  x += a;
  if (x >= MOD) {
    x -= MOD;
  }
}
const int N = 100000;
int mu[N], mpowers[N], minvs[N];
std::vector<int> divs[N], sfdivs[N];
int gsum(int g, int n) { return (1LL + MOD - mpowers[g * n]) * minvs[g] % MOD; }
int main() {
  mu[1] = 1;
  for (int i = 1; i < N; ++i) {
    for (int j = i; j < N; j += i) {
      divs[j].push_back(i);
      if (mu[i]) {
        sfdivs[j].push_back(i);
      }
    }
    for (int j = i + i; j < N; j += i) {
      mu[j] -= mu[i];
    }
  }
  int n, m;
  scanf("%d%d", &n, &m);
  if (m == 1) {
    printf("%d\n", (int)(1LL * (n - 1) * (n - 1) % MOD));
  } else {
    mpowers[0] = 1;
    for (int i = 1; i < n; ++i) {
      mpowers[i] = 1LL * mpowers[i - 1] * m % MOD;
      minvs[i] = 1LL * mpowers[i] * power(1 + MOD - mpowers[i], MOD - 2) % MOD;
    }
    int inv_kn = power(m, 1LL * n * (MOD - 2) % (MOD - 1));
    int result = 1LL * power(gsum(1, n - 1), 2) * inv_kn % MOD;
    for (int p = 1; p < n; ++p) {
      for (int g : divs[p]) {
        int le_count = 0, le_sum = 0;
        for (int _ : sfdivs[p / g]) {
          int gg = g * _;
          int count = std::min(n - 1, n + g - p) / gg;
          le_count += mu[_] * count;
          int sum = gsum(gg, count);
          add(le_sum, mu[_] > 0 ? sum : MOD - sum);
        }
        add(result, 1LL * le_count * mpowers[g] % MOD);
        add(result, MOD - 1LL * le_sum * mpowers[p] % MOD * inv_kn % MOD);
      }
    }
    printf("%d\n", (int)(1LL * result * inv_kn % MOD));
  }
}

```