#include <bits/stdc++.h>
using namespace std;
long long n, m, pim[200005], sum[200005], sta[200005], mi[200005], num[200005],
    ans, phi[200005], top;
long long add(long long x) { return x >= 1000000007 ? x - 1000000007 : x; }
long long jian(long long x) { return x < 0 ? x + 1000000007 : x; }
void prepare(long long n) {
  for (long long i = 2; i <= n; i++) {
    if (!pim[i]) {
      phi[i] = i - 1;
      sta[++top] = i;
    }
    for (long long t = 1; t <= top; t++) {
      if (i * sta[t] > n) break;
      pim[i * sta[t]] = 1;
      if (i % sta[t] == 0) {
        phi[i * sta[t]] = phi[i] * sta[t] % 1000000007;
        break;
      } else {
        phi[i * sta[t]] = phi[i] * (sta[t] - 1) % 1000000007;
      }
    }
  }
  for (long long i = 1; i <= n; i++) {
    sum[i] = add(phi[i] + sum[i - 1]);
  }
}
long long ksm(long long x, long long y) {
  long long ans = 1, t = x;
  while (y) {
    if (y & 1) ans = ans * t % 1000000007;
    t = t * t % 1000000007;
    y >>= 1;
  }
  return ans;
}
int main() {
  scanf("%lld%lld", &n, &m);
  prepare(n);
  mi[0] = 1;
  for (long long i = 1; i <= n; i++) mi[i] = mi[i - 1] * m % 1000000007;
  for (long long s = n + 1; s <= 2 * n - 2; s++) {
    ans = (ans + mi[s - n] * (2 * n - s - 1)) % 1000000007;
  }
  for (long long p = 1; p < n; p++) {
    num[p] = add(num[p] + sum[(n - 1) / p + 1]);
    num[((n - 1) / p + 1) * p - n] =
        jian(num[((n - 1) / p + 1) * p - n] - phi[(n - 1) / p + 1]);
  }
  for (long long i = 1; i <= n; i++) ans = (ans + num[i] * mi[i]) % 1000000007;
  printf("%lld", ans * ksm(mi[n], 1000000007 - 2) % 1000000007);
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 n, m, pim[200005], sum[200005], sta[200005], mi[200005], num[200005],
    ans, phi[200005], top;
long long add(long long x) { return x >= 1000000007 ? x - 1000000007 : x; }
long long jian(long long x) { return x < 0 ? x + 1000000007 : x; }
void prepare(long long n) {
  for (long long i = 2; i <= n; i++) {
    if (!pim[i]) {
      phi[i] = i - 1;
      sta[++top] = i;
    }
    for (long long t = 1; t <= top; t++) {
      if (i * sta[t] > n) break;
      pim[i * sta[t]] = 1;
      if (i % sta[t] == 0) {
        phi[i * sta[t]] = phi[i] * sta[t] % 1000000007;
        break;
      } else {
        phi[i * sta[t]] = phi[i] * (sta[t] - 1) % 1000000007;
      }
    }
  }
  for (long long i = 1; i <= n; i++) {
    sum[i] = add(phi[i] + sum[i - 1]);
  }
}
long long ksm(long long x, long long y) {
  long long ans = 1, t = x;
  while (y) {
    if (y & 1) ans = ans * t % 1000000007;
    t = t * t % 1000000007;
    y >>= 1;
  }
  return ans;
}
int main() {
  scanf("%lld%lld", &n, &m);
  prepare(n);
  mi[0] = 1;
  for (long long i = 1; i <= n; i++) mi[i] = mi[i - 1] * m % 1000000007;
  for (long long s = n + 1; s <= 2 * n - 2; s++) {
    ans = (ans + mi[s - n] * (2 * n - s - 1)) % 1000000007;
  }
  for (long long p = 1; p < n; p++) {
    num[p] = add(num[p] + sum[(n - 1) / p + 1]);
    num[((n - 1) / p + 1) * p - n] =
        jian(num[((n - 1) / p + 1) * p - n] - phi[(n - 1) / p + 1]);
  }
  for (long long i = 1; i <= n; i++) ans = (ans + num[i] * mi[i]) % 1000000007;
  printf("%lld", ans * ksm(mi[n], 1000000007 - 2) % 1000000007);
}

```