#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
long long int powmod(long long int a, long long int k) {
  long long int ap = a, ans = 1;
  while (k) {
    if (k & 1) {
      ans *= ap;
      ans %= MOD;
    }
    ap = ap * ap;
    ap %= MOD;
    k >>= 1;
  }
  return ans;
}
long long int inv(long long int a) { return powmod(a, MOD - 2); }
int main() {
  int n, k;
  cin >> n >> k;
  long long int ans = 0;
  long long int kp[200020];
  kp[0] = 1;
  for (int i = 1; i <= 2 * n; i++) kp[i] = kp[i - 1] * k % MOD;
  vector<int> vd[200010];
  for (int i = 2; i <= 2 * n - 2; i++) {
    for (int j = 1; j * j <= i; j++) {
      if (i % j == 0) {
        vd[i].push_back(j);
        if (j * j < i) vd[i].push_back(i / j);
      }
    }
    sort(vd[i].begin(), vd[i].end());
  }
  for (int m = 2; m <= 2 * n - 2; m++) {
    vector<int> cnt(vd[m].size());
    for (int i = (int)vd[m].size() - 1; i >= 0; i--) {
      int d = vd[m][i];
      if (m <= n)
        cnt[i] = (m - 1) / d;
      else
        cnt[i] = (n - 1) / d - (m - n) / d;
      for (int j = i + 1; j < vd[m].size(); j++) {
        if (vd[m][j] % d == 0) cnt[i] -= cnt[j];
      }
    }
    for (int i = 0; i < vd[m].size(); i++) {
      (ans += cnt[i] * kp[max(vd[m][i], m - n)]) %= MOD;
    }
  }
  (ans *= inv(kp[n])) %= MOD;
  cout << ans << endl;
  return 0;
}

### 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>
using namespace std;
const long long int MOD = 1e9 + 7;
long long int powmod(long long int a, long long int k) {
  long long int ap = a, ans = 1;
  while (k) {
    if (k & 1) {
      ans *= ap;
      ans %= MOD;
    }
    ap = ap * ap;
    ap %= MOD;
    k >>= 1;
  }
  return ans;
}
long long int inv(long long int a) { return powmod(a, MOD - 2); }
int main() {
  int n, k;
  cin >> n >> k;
  long long int ans = 0;
  long long int kp[200020];
  kp[0] = 1;
  for (int i = 1; i <= 2 * n; i++) kp[i] = kp[i - 1] * k % MOD;
  vector<int> vd[200010];
  for (int i = 2; i <= 2 * n - 2; i++) {
    for (int j = 1; j * j <= i; j++) {
      if (i % j == 0) {
        vd[i].push_back(j);
        if (j * j < i) vd[i].push_back(i / j);
      }
    }
    sort(vd[i].begin(), vd[i].end());
  }
  for (int m = 2; m <= 2 * n - 2; m++) {
    vector<int> cnt(vd[m].size());
    for (int i = (int)vd[m].size() - 1; i >= 0; i--) {
      int d = vd[m][i];
      if (m <= n)
        cnt[i] = (m - 1) / d;
      else
        cnt[i] = (n - 1) / d - (m - n) / d;
      for (int j = i + 1; j < vd[m].size(); j++) {
        if (vd[m][j] % d == 0) cnt[i] -= cnt[j];
      }
    }
    for (int i = 0; i < vd[m].size(); i++) {
      (ans += cnt[i] * kp[max(vd[m][i], m - n)]) %= MOD;
    }
  }
  (ans *= inv(kp[n])) %= MOD;
  cout << ans << endl;
  return 0;
}

```