#include <bits/stdc++.h>
using namespace std;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = 0;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f ? x : -x;
}
const int maxn = 2e5 + 10, mod = 1e9 + 7;
int bin[maxn], pri[maxn], vis[maxn], tot, mu[maxn], mus[maxn], Ans[maxn];
int Pow(int x, int p) {
  int r = 1;
  while (p) {
    if (p & 1) r = (long long)x * r % mod;
    x = (long long)x * x % mod;
    p >>= 1;
  }
  return r;
}
int main() {
  int n = read(), k = read();
  bin[0] = 1;
  for (int i = 1; i <= n; ++i) bin[i] = (long long)bin[i - 1] * k % mod;
  mu[1] = 1;
  for (int i = 2; i <= (n << 1); ++i) {
    if (!vis[i]) {
      mu[i] = i - 1;
      pri[++tot] = i;
    }
    for (int j = 1; j <= tot && pri[j] * i <= (n << 1); ++j) {
      int tmp = pri[j] * i;
      vis[tmp] = 1;
      if (i % pri[j] == 0) {
        mu[tmp] = (long long)mu[i] * pri[j] % mod;
        break;
      }
      mu[tmp] = (long long)mu[i] * mu[pri[j]] % mod;
    }
  }
  for (int i = 2; i <= (n << 1); ++i) (mus[i] = mus[i - 1] + mu[i]) %= mod;
  int ans = 0;
  for (int i = 1; i < n - 1; ++i) Ans[i] = (n - 1 - i);
  for (int i = 1; i < n; ++i) {
    int d = (n + i - 1) / i;
    (ans += (long long)bin[i] * mus[d] % mod) %= mod;
    (Ans[d * i - n] -= mu[d]) %= mod;
  }
  for (int i = 1; i < n - 1; ++i)
    (ans += (long long)bin[i] * Ans[i] % mod) %= mod;
  cout << (long long)ans * Pow(Pow(k, mod - 2), n) % mod << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = 0;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f ? x : -x;
}
const int maxn = 2e5 + 10, mod = 1e9 + 7;
int bin[maxn], pri[maxn], vis[maxn], tot, mu[maxn], mus[maxn], Ans[maxn];
int Pow(int x, int p) {
  int r = 1;
  while (p) {
    if (p & 1) r = (long long)x * r % mod;
    x = (long long)x * x % mod;
    p >>= 1;
  }
  return r;
}
int main() {
  int n = read(), k = read();
  bin[0] = 1;
  for (int i = 1; i <= n; ++i) bin[i] = (long long)bin[i - 1] * k % mod;
  mu[1] = 1;
  for (int i = 2; i <= (n << 1); ++i) {
    if (!vis[i]) {
      mu[i] = i - 1;
      pri[++tot] = i;
    }
    for (int j = 1; j <= tot && pri[j] * i <= (n << 1); ++j) {
      int tmp = pri[j] * i;
      vis[tmp] = 1;
      if (i % pri[j] == 0) {
        mu[tmp] = (long long)mu[i] * pri[j] % mod;
        break;
      }
      mu[tmp] = (long long)mu[i] * mu[pri[j]] % mod;
    }
  }
  for (int i = 2; i <= (n << 1); ++i) (mus[i] = mus[i - 1] + mu[i]) %= mod;
  int ans = 0;
  for (int i = 1; i < n - 1; ++i) Ans[i] = (n - 1 - i);
  for (int i = 1; i < n; ++i) {
    int d = (n + i - 1) / i;
    (ans += (long long)bin[i] * mus[d] % mod) %= mod;
    (Ans[d * i - n] -= mu[d]) %= mod;
  }
  for (int i = 1; i < n - 1; ++i)
    (ans += (long long)bin[i] * Ans[i] % mod) %= mod;
  cout << (long long)ans * Pow(Pow(k, mod - 2), n) % mod << endl;
  return 0;
}

```