#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int w = 0, x = 0;
  char c = getchar();
  while (!isdigit(c)) w |= c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();
  return w ? -x : x;
}
namespace star {
const int maxn = 1e5 + 10, mod = 1e9 + 7;
int n, k, ans, mul[maxn], phi[maxn], p[maxn], tot;
bool mark[maxn];
inline int fpow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
inline void work() {
  n = read(), k = read();
  mul[0] = 1;
  for (int i = 1; i <= n; i++) mul[i] = 1ll * mul[i - 1] * k % mod;
  mul[0] = 0;
  for (int i = 1, a = n - 2; i <= n and a > 0; i++, a--)
    ans = (ans + 1ll * a * mul[i]) % mod;
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!mark[i]) p[++tot] = i, phi[i] = i - 1;
    for (int j = 1, tmp; j <= tot and (tmp = i * p[j]) <= n; j++) {
      mark[tmp] = true;
      if (i % p[j] == 0) {
        phi[tmp] = phi[i] * p[j];
        break;
      }
      phi[tmp] = phi[i] * phi[p[j]];
    }
  }
  for (int i = 2; i <= n; i++) phi[i] = (phi[i] + phi[i - 1]) % mod;
  for (int x, i = 1; i < n; i++)
    x = (n - 1) / i + 1,
    ans = (ans - 1ll * (phi[x] - phi[x - 1] + mod) * mul[x * i - n] % mod +
           (phi[x] - 1ll + mod) * mul[i] % mod + mod) %
          mod;
  printf("%lld\n", 1ll * ans * fpow(mul[n], mod - 2) % mod);
}
}  // namespace star
signed main() {
  star::work();
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
inline int read() {
  int w = 0, x = 0;
  char c = getchar();
  while (!isdigit(c)) w |= c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();
  return w ? -x : x;
}
namespace star {
const int maxn = 1e5 + 10, mod = 1e9 + 7;
int n, k, ans, mul[maxn], phi[maxn], p[maxn], tot;
bool mark[maxn];
inline int fpow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
inline void work() {
  n = read(), k = read();
  mul[0] = 1;
  for (int i = 1; i <= n; i++) mul[i] = 1ll * mul[i - 1] * k % mod;
  mul[0] = 0;
  for (int i = 1, a = n - 2; i <= n and a > 0; i++, a--)
    ans = (ans + 1ll * a * mul[i]) % mod;
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!mark[i]) p[++tot] = i, phi[i] = i - 1;
    for (int j = 1, tmp; j <= tot and (tmp = i * p[j]) <= n; j++) {
      mark[tmp] = true;
      if (i % p[j] == 0) {
        phi[tmp] = phi[i] * p[j];
        break;
      }
      phi[tmp] = phi[i] * phi[p[j]];
    }
  }
  for (int i = 2; i <= n; i++) phi[i] = (phi[i] + phi[i - 1]) % mod;
  for (int x, i = 1; i < n; i++)
    x = (n - 1) / i + 1,
    ans = (ans - 1ll * (phi[x] - phi[x - 1] + mod) * mul[x * i - n] % mod +
           (phi[x] - 1ll + mod) * mul[i] % mod + mod) %
          mod;
  printf("%lld\n", 1ll * ans * fpow(mul[n], mod - 2) % mod);
}
}  // namespace star
signed main() {
  star::work();
  return 0;
}

```