#include <bits/stdc++.h>
inline int read() {
  int data = 0, w = 1;
  char ch = getchar();
  while (ch != '-' && !std::isdigit(ch)) ch = std::getchar();
  if (ch == '-') w = -1, ch = std::getchar();
  while (std::isdigit(ch)) data = data * 10 + (ch ^ 48), ch = std::getchar();
  return data * w;
}
constexpr int N(1e5 + 10), Mod(1e9 + 7);
inline int upd(const int &x) { return x + (x >> 31 & Mod); }
int fastpow(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = 1ll * x * x % Mod)
    if (y & 1) ans = 1ll * ans * x % Mod;
  return ans;
}
int n, K, ans, Pow[N << 1], invK, sum[N], f[N], g[N], mu[N];
int main() {
  n = read(), K = read(), Pow[0] = mu[1] = 1,
  invK = fastpow(fastpow(K, n), Mod - 2);
  for (int i = 1; i <= (n << 1); i++) Pow[i] = 1ll * Pow[i - 1] * K % Mod;
  for (int i = 2; i <= n + 1; i++)
    sum[i] = (sum[i - 1] + 1ll * Pow[i] * (i - 1)) % Mod;
  for (int i = 2; i <= n + 1; i++) f[i] = upd(f[i - 1] + i - 1 - Mod);
  ans = 1ll * Pow[n] * (n - 1) % Mod;
  for (int i = 2; i < n; i++)
    ans = (ans + 1ll * (Pow[i] + Pow[(n << 1) - i]) * (i - 1)) % Mod;
  for (int i = 1; i <= n; i++)
    for (int j = i + i; j <= n; j += i) mu[j] -= mu[i];
  for (int i = 1; i <= n; i++)
    for (int j = (n + i - 1) / i, k = 2; k <= j; k++)
      g[i] = (g[i] + 1ll * Pow[i * k] * (k - 1)) % Mod;
  for (int i = 1; i < n; i++)
    for (int j = 1, t; (t = i * j) < n; j++) {
      int t1 = (n + i - 1) / t, t2 = (n + t - 1) / t, c = g[t];
      if (t2 > t1) c = (c - 1ll * Pow[t2 * t] * (t2 - 1)) % Mod;
      ans = (ans - 1ll * c * mu[j]) % Mod;
    }
  ans = 1ll * ans * invK % Mod;
  for (int i = 1; i < n; i++)
    for (int j = 1, t; (t = i * j) < n; j++)
      ans = (ans + 1ll * f[(n + i - 1) / t] * Pow[i] * mu[j]) % Mod;
  printf("%lld\n", 1ll * upd(ans) * invK % Mod);
  return 0;
}

### Prompt

Please formulate 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>
inline int read() {
  int data = 0, w = 1;
  char ch = getchar();
  while (ch != '-' && !std::isdigit(ch)) ch = std::getchar();
  if (ch == '-') w = -1, ch = std::getchar();
  while (std::isdigit(ch)) data = data * 10 + (ch ^ 48), ch = std::getchar();
  return data * w;
}
constexpr int N(1e5 + 10), Mod(1e9 + 7);
inline int upd(const int &x) { return x + (x >> 31 & Mod); }
int fastpow(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = 1ll * x * x % Mod)
    if (y & 1) ans = 1ll * ans * x % Mod;
  return ans;
}
int n, K, ans, Pow[N << 1], invK, sum[N], f[N], g[N], mu[N];
int main() {
  n = read(), K = read(), Pow[0] = mu[1] = 1,
  invK = fastpow(fastpow(K, n), Mod - 2);
  for (int i = 1; i <= (n << 1); i++) Pow[i] = 1ll * Pow[i - 1] * K % Mod;
  for (int i = 2; i <= n + 1; i++)
    sum[i] = (sum[i - 1] + 1ll * Pow[i] * (i - 1)) % Mod;
  for (int i = 2; i <= n + 1; i++) f[i] = upd(f[i - 1] + i - 1 - Mod);
  ans = 1ll * Pow[n] * (n - 1) % Mod;
  for (int i = 2; i < n; i++)
    ans = (ans + 1ll * (Pow[i] + Pow[(n << 1) - i]) * (i - 1)) % Mod;
  for (int i = 1; i <= n; i++)
    for (int j = i + i; j <= n; j += i) mu[j] -= mu[i];
  for (int i = 1; i <= n; i++)
    for (int j = (n + i - 1) / i, k = 2; k <= j; k++)
      g[i] = (g[i] + 1ll * Pow[i * k] * (k - 1)) % Mod;
  for (int i = 1; i < n; i++)
    for (int j = 1, t; (t = i * j) < n; j++) {
      int t1 = (n + i - 1) / t, t2 = (n + t - 1) / t, c = g[t];
      if (t2 > t1) c = (c - 1ll * Pow[t2 * t] * (t2 - 1)) % Mod;
      ans = (ans - 1ll * c * mu[j]) % Mod;
    }
  ans = 1ll * ans * invK % Mod;
  for (int i = 1; i < n; i++)
    for (int j = 1, t; (t = i * j) < n; j++)
      ans = (ans + 1ll * f[(n + i - 1) / t] * Pow[i] * mu[j]) % Mod;
  printf("%lld\n", 1ll * upd(ans) * invK % Mod);
  return 0;
}

```