#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool flag = false;
  while (!isdigit(c)) {
    if (c == '-') flag = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  if (flag) x = -x;
}
int n, k, tot;
long long ans;
int pri[200010];
long long mu[200010], pw[200010];
bool tag[200010];
vector<int> ve[200010];
long long inv(long long x) {
  long long v = 1, y = 1000000007 - 2;
  while (y) {
    if (y & 1) v = v * x % 1000000007;
    x = x * x % 1000000007, y >>= 1;
  }
  return v;
}
void init(int n) {
  mu[1] = pw[0] = 1;
  for (int i = 1; i <= n; ++i) pw[i] = pw[i - 1] * k % 1000000007;
  for (int i = 2; i <= n; ++i) {
    if (!tag[i]) mu[pri[++tot] = i] = 1000000007 - 1;
    for (int j = 1; j <= tot; ++j) {
      int k = i * pri[j];
      if (k > n) break;
      tag[k] = true;
      if (i % pri[j])
        mu[k] = 1000000007 - mu[i];
      else
        break;
    }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; j += i) ve[j].push_back(i);
}
long long calc(int lim, int sum) {
  if (lim <= 0 || sum <= 1) return 0;
  long long l = max(1, sum - lim), r = min(lim, sum - 1), v = 0;
  if (l > r) return 0;
  for (int i = 0; i < ve[sum].size(); ++i) {
    int d = ve[sum][i];
    v = (v + mu[d] * (r / d - (l - 1) / d) % 1000000007) % 1000000007;
  }
  return v;
}
int main() {
  read(n), read(k), init(2 * n);
  for (int s = 2; s <= 2 * n - 2; ++s) {
    for (int i = 0; i < ve[s].size(); ++i) {
      int g = ve[s][i];
      ans = (ans + calc((n - 1) / g, s / g) * pw[max(s - n, g)] % 1000000007) %
            1000000007;
    }
  }
  printf("%lld", ans * inv(pw[n]) % 1000000007);
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool flag = false;
  while (!isdigit(c)) {
    if (c == '-') flag = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  if (flag) x = -x;
}
int n, k, tot;
long long ans;
int pri[200010];
long long mu[200010], pw[200010];
bool tag[200010];
vector<int> ve[200010];
long long inv(long long x) {
  long long v = 1, y = 1000000007 - 2;
  while (y) {
    if (y & 1) v = v * x % 1000000007;
    x = x * x % 1000000007, y >>= 1;
  }
  return v;
}
void init(int n) {
  mu[1] = pw[0] = 1;
  for (int i = 1; i <= n; ++i) pw[i] = pw[i - 1] * k % 1000000007;
  for (int i = 2; i <= n; ++i) {
    if (!tag[i]) mu[pri[++tot] = i] = 1000000007 - 1;
    for (int j = 1; j <= tot; ++j) {
      int k = i * pri[j];
      if (k > n) break;
      tag[k] = true;
      if (i % pri[j])
        mu[k] = 1000000007 - mu[i];
      else
        break;
    }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; j += i) ve[j].push_back(i);
}
long long calc(int lim, int sum) {
  if (lim <= 0 || sum <= 1) return 0;
  long long l = max(1, sum - lim), r = min(lim, sum - 1), v = 0;
  if (l > r) return 0;
  for (int i = 0; i < ve[sum].size(); ++i) {
    int d = ve[sum][i];
    v = (v + mu[d] * (r / d - (l - 1) / d) % 1000000007) % 1000000007;
  }
  return v;
}
int main() {
  read(n), read(k), init(2 * n);
  for (int s = 2; s <= 2 * n - 2; ++s) {
    for (int i = 0; i < ve[s].size(); ++i) {
      int g = ve[s][i];
      ans = (ans + calc((n - 1) / g, s / g) * pw[max(s - n, g)] % 1000000007) %
            1000000007;
    }
  }
  printf("%lld", ans * inv(pw[n]) % 1000000007);
  return 0;
}

```