#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int power(int x, int t) {
  int ret = 1;
  while (t > 0) {
    if (t & 1) ret = 1LL * ret * x % MOD;
    x = 1LL * x * x % MOD;
    t >>= 1;
  }
  return ret;
}
int mu[200005];
int cnt[200005];
int getcnt(int sumc, int lima) { return min(sumc - 1, lima * 2 - sumc + 1); }
int solve(int k, int coef, int lim, int sdiv) {
  for (int i = 1; i <= (int)(2 * lim); i++) cnt[i] = 0;
  for (int i = 1; i <= (int)(lim); i++)
    for (int j = i; j <= 2 * lim; j += i)
      cnt[j] = (cnt[j] + 1LL * (MOD + mu[i]) * getcnt(j / i, lim / i)) % MOD;
  int ret = 0, cpow = 1;
  for (int i = 1; i <= (int)(2 * lim); i++) {
    cpow = 1LL * cpow * k % MOD;
    if (i > sdiv)
      ret = (ret + 1LL * cpow * coef % MOD * cnt[i]) % MOD;
    else
      ret = (ret + 1LL * k * cnt[i]) % MOD;
  }
  return ret;
}
int main() {
  mu[1] = 1;
  for (int i = 1; i <= (int)(200000); i++)
    for (int j = i * 2; j <= 200000; j += i) mu[j] -= mu[i];
  int n, k;
  scanf("%d%d", &n, &k);
  int ans = 0;
  for (int t = 1; t <= (int)(n); t++)
    ans = (ans +
           solve(power(k, t), power(k, MOD - 1 - n), (n - 1) / t, n / t + 1)) %
          MOD;
  ans = 1LL * ans * power(k, MOD - 1 - n) % MOD;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
const int MOD = 1e9 + 7;
int power(int x, int t) {
  int ret = 1;
  while (t > 0) {
    if (t & 1) ret = 1LL * ret * x % MOD;
    x = 1LL * x * x % MOD;
    t >>= 1;
  }
  return ret;
}
int mu[200005];
int cnt[200005];
int getcnt(int sumc, int lima) { return min(sumc - 1, lima * 2 - sumc + 1); }
int solve(int k, int coef, int lim, int sdiv) {
  for (int i = 1; i <= (int)(2 * lim); i++) cnt[i] = 0;
  for (int i = 1; i <= (int)(lim); i++)
    for (int j = i; j <= 2 * lim; j += i)
      cnt[j] = (cnt[j] + 1LL * (MOD + mu[i]) * getcnt(j / i, lim / i)) % MOD;
  int ret = 0, cpow = 1;
  for (int i = 1; i <= (int)(2 * lim); i++) {
    cpow = 1LL * cpow * k % MOD;
    if (i > sdiv)
      ret = (ret + 1LL * cpow * coef % MOD * cnt[i]) % MOD;
    else
      ret = (ret + 1LL * k * cnt[i]) % MOD;
  }
  return ret;
}
int main() {
  mu[1] = 1;
  for (int i = 1; i <= (int)(200000); i++)
    for (int j = i * 2; j <= 200000; j += i) mu[j] -= mu[i];
  int n, k;
  scanf("%d%d", &n, &k);
  int ans = 0;
  for (int t = 1; t <= (int)(n); t++)
    ans = (ans +
           solve(power(k, t), power(k, MOD - 1 - n), (n - 1) / t, n / t + 1)) %
          MOD;
  ans = 1LL * ans * power(k, MOD - 1 - n) % MOD;
  printf("%d\n", ans);
  return 0;
}

```