#include <bits/stdc++.h>
const int mod = 1e9 + 7;
using namespace std;
int T, n, m, p = 0, phi[100010], S[100010], pr[100010], ans[100010];
bool isnt[100010];
int& qmo(int& x) { return x += (x >> 31) & mod; }
int ksm(int a, int k) {
  int res = 1;
  for (; k; k >>= 1, a = 1ll * a * a % mod)
    if (k & 1) res = 1ll * res * a % mod;
  return res;
}
void init(int N) {
  phi[1] = S[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (!isnt[i]) {
      pr[++p] = i;
      phi[i] = i - 1;
    }
    for (int j = 1; j <= p && pr[j] * i <= N; j++) {
      isnt[pr[j] * i] = 1;
      if (i % pr[j] == 0) {
        phi[pr[j] * i] = phi[i] * pr[j];
        break;
      } else
        phi[pr[j] * i] = phi[i] * phi[pr[j]];
    }
    qmo(S[i] = S[i - 1] + phi[i] - mod);
  }
}
int main() {
  scanf("%d%d", &n, &m), init(n);
  for (int i = 1; i < n; i++) {
    int t = (n + i - 1) / i;
    qmo(qmo(qmo(ans[i] += S[t] - mod) -= 1) += (n - i - 1) - mod);
    if (n % i) qmo(ans[t * i - n] -= phi[t]);
  }
  int res = 0;
  for (int i = 0, nw = 1; i < n; i++)
    res = (res + 1ll * nw * ans[i]) % mod, nw = 1ll * nw * m % mod;
  res = 1ll * res * ksm(ksm(m, n), mod - 2) % mod;
  printf("%d\n", res);
}

### Prompt

Your task is to create 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>
const int mod = 1e9 + 7;
using namespace std;
int T, n, m, p = 0, phi[100010], S[100010], pr[100010], ans[100010];
bool isnt[100010];
int& qmo(int& x) { return x += (x >> 31) & mod; }
int ksm(int a, int k) {
  int res = 1;
  for (; k; k >>= 1, a = 1ll * a * a % mod)
    if (k & 1) res = 1ll * res * a % mod;
  return res;
}
void init(int N) {
  phi[1] = S[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (!isnt[i]) {
      pr[++p] = i;
      phi[i] = i - 1;
    }
    for (int j = 1; j <= p && pr[j] * i <= N; j++) {
      isnt[pr[j] * i] = 1;
      if (i % pr[j] == 0) {
        phi[pr[j] * i] = phi[i] * pr[j];
        break;
      } else
        phi[pr[j] * i] = phi[i] * phi[pr[j]];
    }
    qmo(S[i] = S[i - 1] + phi[i] - mod);
  }
}
int main() {
  scanf("%d%d", &n, &m), init(n);
  for (int i = 1; i < n; i++) {
    int t = (n + i - 1) / i;
    qmo(qmo(qmo(ans[i] += S[t] - mod) -= 1) += (n - i - 1) - mod);
    if (n % i) qmo(ans[t * i - n] -= phi[t]);
  }
  int res = 0;
  for (int i = 0, nw = 1; i < n; i++)
    res = (res + 1ll * nw * ans[i]) % mod, nw = 1ll * nw * m % mod;
  res = 1ll * res * ksm(ksm(m, n), mod - 2) % mod;
  printf("%d\n", res);
}

```