#include <bits/stdc++.h>
inline int fmo(int x) { return x + ((x >> 31) & 1000000007); }
inline int fp(int x, int k = 1000000007 - 2) {
  int res = 1;
  for (; k; k >>= 1, x = 1ll * x * x % 1000000007)
    if (k & 1) res = 1ll * res * x % 1000000007;
  return res;
}
int pr[100005], _pr, mu[100005];
bool npr[100005];
inline void si(int x) {
  mu[1] = 1;
  for (int i = 2; i <= x; i++) {
    if (!npr[i]) {
      pr[++_pr] = i;
      mu[i] = 1000000007 - 1;
    }
    for (int j = 1; j <= _pr && i * pr[j] <= x; j++) {
      npr[i * pr[j]] = 1;
      if (i % pr[j] == 0) break;
      mu[i * pr[j]] = fmo(-mu[i]);
    }
  }
}
int n, k;
int pw[100005];
int ans;
int main() {
  scanf("%d%d", &n, &k);
  si(n);
  pw[0] = 1;
  for (int i = 1; i <= n; i++) pw[i] = 1ll * pw[i - 1] * k % 1000000007;
  for (int d = 1; d < n; d++)
    for (int t = 1; t <= (n - 1) / d; t++) {
      int x = (n - 1) / d / t, tmp = 0;
      for (int i = 2; i <= 2 * x; i++)
        tmp = fmo(tmp +
                  1ll * (x - std::abs(x + 1 - i)) *
                      pw[std::max(i * d * t - n, d)] % 1000000007 -
                  1000000007);
      ans = fmo(ans + 1ll * mu[t] * tmp % 1000000007 - 1000000007);
    }
  ans = 1ll * ans * fp(pw[n]) % 1000000007;
  printf("%d\n", ans);
}

### Prompt

Your task is to create 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>
inline int fmo(int x) { return x + ((x >> 31) & 1000000007); }
inline int fp(int x, int k = 1000000007 - 2) {
  int res = 1;
  for (; k; k >>= 1, x = 1ll * x * x % 1000000007)
    if (k & 1) res = 1ll * res * x % 1000000007;
  return res;
}
int pr[100005], _pr, mu[100005];
bool npr[100005];
inline void si(int x) {
  mu[1] = 1;
  for (int i = 2; i <= x; i++) {
    if (!npr[i]) {
      pr[++_pr] = i;
      mu[i] = 1000000007 - 1;
    }
    for (int j = 1; j <= _pr && i * pr[j] <= x; j++) {
      npr[i * pr[j]] = 1;
      if (i % pr[j] == 0) break;
      mu[i * pr[j]] = fmo(-mu[i]);
    }
  }
}
int n, k;
int pw[100005];
int ans;
int main() {
  scanf("%d%d", &n, &k);
  si(n);
  pw[0] = 1;
  for (int i = 1; i <= n; i++) pw[i] = 1ll * pw[i - 1] * k % 1000000007;
  for (int d = 1; d < n; d++)
    for (int t = 1; t <= (n - 1) / d; t++) {
      int x = (n - 1) / d / t, tmp = 0;
      for (int i = 2; i <= 2 * x; i++)
        tmp = fmo(tmp +
                  1ll * (x - std::abs(x + 1 - i)) *
                      pw[std::max(i * d * t - n, d)] % 1000000007 -
                  1000000007);
      ans = fmo(ans + 1ll * mu[t] * tmp % 1000000007 - 1000000007);
    }
  ans = 1ll * ans * fp(pw[n]) % 1000000007;
  printf("%d\n", ans);
}

```