#include <bits/stdc++.h>
using namespace std;
const int Mod = 1e9 + 7;
int add(int a, int b) { return a + b >= Mod ? a + b - Mod : a + b; }
int dec(int a, int b) { return a - b < 0 ? a - b + Mod : a - b; }
int mul(int a, int b) { return 1ll * a * b % Mod; }
void Add(int &a, int b) { a = add(a, b); }
void Dec(int &a, int b) { a = dec(a, b); }
void Mul(int &a, int b) { a = mul(a, b); }
int ksm(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, Mul(a, a))
    if (b & 1) Mul(ans, a);
  return ans;
}
const int N = 2e5 + 50;
int n, k, pw[N], mu[N];
vector<int> G[N];
vector<int> prm;
bool ex[N];
int main() {
  cin >> n >> k;
  pw[0] = 1;
  for (int i = 1; i <= n; i++) pw[i] = mul(pw[i - 1], k);
  mu[1] = 1;
  for (int i = 2; i <= n + n; i++) {
    if (!ex[i]) prm.push_back(i), mu[i] = dec(0, 1);
    for (int c : prm) {
      if (c * i > n) break;
      ex[c * i] = true;
      if (i % c == 0) break;
      mu[i * c] = dec(0, mu[i]);
    }
  }
  for (int i = 1; i <= n + n; i++)
    for (int j = i; j <= n + n; j += i) G[j].push_back(i);
  int ans = 0;
  for (int d = 1; d < n; d++)
    for (int t = 1; t <= 2 * ((n - 1) / d); t++) {
      int z = pw[max(d, d * t - n)];
      int c = 0;
      for (int l : G[t])
        if (mu[l]) {
          int lm = (n - 1) / d / l;
          if (lm + lm < t / l) continue;
          Add(c, mul(mu[l], min(lm + lm - t / l + 1, t / l - 1)));
        }
      Add(ans, mul(z, c));
    }
  cout << mul(ans, ksm(pw[n], Mod - 2));
  return 0;
}

### Prompt

Generate 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 int Mod = 1e9 + 7;
int add(int a, int b) { return a + b >= Mod ? a + b - Mod : a + b; }
int dec(int a, int b) { return a - b < 0 ? a - b + Mod : a - b; }
int mul(int a, int b) { return 1ll * a * b % Mod; }
void Add(int &a, int b) { a = add(a, b); }
void Dec(int &a, int b) { a = dec(a, b); }
void Mul(int &a, int b) { a = mul(a, b); }
int ksm(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, Mul(a, a))
    if (b & 1) Mul(ans, a);
  return ans;
}
const int N = 2e5 + 50;
int n, k, pw[N], mu[N];
vector<int> G[N];
vector<int> prm;
bool ex[N];
int main() {
  cin >> n >> k;
  pw[0] = 1;
  for (int i = 1; i <= n; i++) pw[i] = mul(pw[i - 1], k);
  mu[1] = 1;
  for (int i = 2; i <= n + n; i++) {
    if (!ex[i]) prm.push_back(i), mu[i] = dec(0, 1);
    for (int c : prm) {
      if (c * i > n) break;
      ex[c * i] = true;
      if (i % c == 0) break;
      mu[i * c] = dec(0, mu[i]);
    }
  }
  for (int i = 1; i <= n + n; i++)
    for (int j = i; j <= n + n; j += i) G[j].push_back(i);
  int ans = 0;
  for (int d = 1; d < n; d++)
    for (int t = 1; t <= 2 * ((n - 1) / d); t++) {
      int z = pw[max(d, d * t - n)];
      int c = 0;
      for (int l : G[t])
        if (mu[l]) {
          int lm = (n - 1) / d / l;
          if (lm + lm < t / l) continue;
          Add(c, mul(mu[l], min(lm + lm - t / l + 1, t / l - 1)));
        }
      Add(ans, mul(z, c));
    }
  cout << mul(ans, ksm(pw[n], Mod - 2));
  return 0;
}

```