#include <bits/stdc++.h>
using namespace std;
using INT = long long;
vector<int> res[201000];
int w[201000];
int cnt[201000];
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < 201000; i++) {
    for (int j = i; j < 201000; j += i) res[j].push_back(i);
  }
  w[1] = 1, cnt[1] = 0;
  for (int i = 2; i < 201000; i++) {
    int p = res[i][1];
    if (i / p % p == 0)
      w[i] = w[i / p];
    else
      w[i] = w[i / p] * p;
    if (w[i] == i) cnt[i] = cnt[i / p] + 1;
  }
  int ans = 0;
  for (int x = 2; x < 2 * n - 1; x++) {
    for (int d : res[x]) {
      if (d >= n) break;
      int L = max(1, (x - n + d) / d), R = (min(n, x) - 1) / d;
      INT tot = 0;
      for (int v : res[w[x / d]]) {
        if (cnt[v] & 1)
          tot -= R / v - (L - 1) / v;
        else
          tot += R / v - (L - 1) / v;
      }
      tot %= 1000000007;
      ans += tot * power(k, max(x - n, d), 1000000007) % 1000000007;
      ans %= 1000000007;
    }
  }
  ans = (INT)ans * power(power(k, n, 1000000007), 1000000007 - 2, 1000000007) %
        1000000007;
  cout << ans << endl;
}

### Prompt

Create a solution in cpp for 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;
using INT = long long;
vector<int> res[201000];
int w[201000];
int cnt[201000];
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < 201000; i++) {
    for (int j = i; j < 201000; j += i) res[j].push_back(i);
  }
  w[1] = 1, cnt[1] = 0;
  for (int i = 2; i < 201000; i++) {
    int p = res[i][1];
    if (i / p % p == 0)
      w[i] = w[i / p];
    else
      w[i] = w[i / p] * p;
    if (w[i] == i) cnt[i] = cnt[i / p] + 1;
  }
  int ans = 0;
  for (int x = 2; x < 2 * n - 1; x++) {
    for (int d : res[x]) {
      if (d >= n) break;
      int L = max(1, (x - n + d) / d), R = (min(n, x) - 1) / d;
      INT tot = 0;
      for (int v : res[w[x / d]]) {
        if (cnt[v] & 1)
          tot -= R / v - (L - 1) / v;
        else
          tot += R / v - (L - 1) / v;
      }
      tot %= 1000000007;
      ans += tot * power(k, max(x - n, d), 1000000007) % 1000000007;
      ans %= 1000000007;
    }
  }
  ans = (INT)ans * power(power(k, n, 1000000007), 1000000007 - 2, 1000000007) %
        1000000007;
  cout << ans << endl;
}

```