#include <bits/stdc++.h>
using namespace std;
long long pow_mod(long long x, int k) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = ans * x % 1000000007;
    x = x * x % 1000000007;
    k >>= 1;
  }
  return ans;
}
long long powd[100005];
int prime[100005], mu[100005];
bool check[100005];
void pre(int n, int k) {
  powd[0] = 1;
  for (int i = 1; i <= n; i++) powd[i] = powd[i - 1] * k % 1000000007;
  mu[1] = 1;
  int tot = 0;
  for (int i = 2; i <= n; i++) {
    if (!check[i]) {
      prime[++tot] = i;
      mu[i] = 1000000007 - 1;
    }
    for (int j = 1; j <= tot && i * prime[j] <= n; j++) {
      check[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
      mu[i * prime[j]] = (1000000007 - mu[i]) % 1000000007;
    }
  }
}
inline long long S(long long n) { return (n * (n - 1) >> 1) % 1000000007; }
int cnt[100005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  pre(n, k);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int t = n / i + 1;
    for (int j = 1; j <= t; j++)
      if (mu[j]) {
        ans = (ans + powd[i] * mu[j] % 1000000007 * S(t / j)) % 1000000007;
        int p = i * j, r = n / p;
        for (int k = 1; p * k <= 2 * n; k++)
          if (p * k > n + i)
            cnt[p * k - n] =
                (cnt[p * k - n] + (long long)mu[j] * (2 * r - k + 1)) %
                1000000007;
      }
  }
  for (int i = 1; i <= n; i++) ans = (ans + cnt[i] * powd[i]) % 1000000007;
  for (int i = 1; i <= n; i++)
    ans = (ans - 2LL * powd[i] % 1000000007 + 1000000007) % 1000000007;
  ans = (ans + powd[n]) % 1000000007;
  printf("%lld\n", ans * pow_mod(powd[n], 1000000007 - 2) % 1000000007);
  return 0;
}

### Prompt

Create a solution in CPP for 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;
long long pow_mod(long long x, int k) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = ans * x % 1000000007;
    x = x * x % 1000000007;
    k >>= 1;
  }
  return ans;
}
long long powd[100005];
int prime[100005], mu[100005];
bool check[100005];
void pre(int n, int k) {
  powd[0] = 1;
  for (int i = 1; i <= n; i++) powd[i] = powd[i - 1] * k % 1000000007;
  mu[1] = 1;
  int tot = 0;
  for (int i = 2; i <= n; i++) {
    if (!check[i]) {
      prime[++tot] = i;
      mu[i] = 1000000007 - 1;
    }
    for (int j = 1; j <= tot && i * prime[j] <= n; j++) {
      check[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
      mu[i * prime[j]] = (1000000007 - mu[i]) % 1000000007;
    }
  }
}
inline long long S(long long n) { return (n * (n - 1) >> 1) % 1000000007; }
int cnt[100005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  pre(n, k);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int t = n / i + 1;
    for (int j = 1; j <= t; j++)
      if (mu[j]) {
        ans = (ans + powd[i] * mu[j] % 1000000007 * S(t / j)) % 1000000007;
        int p = i * j, r = n / p;
        for (int k = 1; p * k <= 2 * n; k++)
          if (p * k > n + i)
            cnt[p * k - n] =
                (cnt[p * k - n] + (long long)mu[j] * (2 * r - k + 1)) %
                1000000007;
      }
  }
  for (int i = 1; i <= n; i++) ans = (ans + cnt[i] * powd[i]) % 1000000007;
  for (int i = 1; i <= n; i++)
    ans = (ans - 2LL * powd[i] % 1000000007 + 1000000007) % 1000000007;
  ans = (ans + powd[n]) % 1000000007;
  printf("%lld\n", ans * pow_mod(powd[n], 1000000007 - 2) % 1000000007);
  return 0;
}

```