#include <bits/stdc++.h>
const long long MOD = 1000000007;
int N, K;
long long O;
int m[100001], pr[100001], v, u[100001], c[100001];
std::vector<int> d[200001];
long long POWER(long long a, long long b) {
  long long r = 1;
  for (; b; b >>= 1) {
    if (b & 1) r = r * a % MOD;
    a = a * a % MOD;
  }
  return r;
}
int main() {
  scanf("%d%d", &N, &K);
  u[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (!m[i]) {
      m[i] = pr[++v] = i;
      u[i] = -1;
    }
    for (int j = 1; j <= v && pr[j] * i <= N && pr[j] <= m[i]; j++) {
      m[pr[j] * i] = pr[j];
      u[pr[j] * i] = pr[j] == m[i] ? 0 : -u[i];
    }
  }
  for (int i = 1; i < N + N; i++)
    for (int j = i; j < N + N; j += i) d[j].push_back(i);
  for (int i = 1; i < N; i++) {
    long long t = 0;
    for (int j = 1; j <= (N - 1) / i; j++)
      if (u[j]) {
        int e = (N - 1) / i / j, r = (N + i) / i / j;
        t += u[j] *
             ((long long)e * e - (long long)(e + e - r) * (e + e - r + 1) / 2);
      }
    std::vector<int> &V = d[N + i];
    int S = V.size();
    for (int j = 0; j < S; j++) c[j] = (N - 1) / V[j] - i / V[j];
    for (int j = S - 1; j >= 0; j--)
      for (int k = j + 1; k < S; k++)
        if (V[k] % V[j] == 0) c[j] -= c[k];
    for (int j = 0; j < S; j++)
      if (V[j] < i) t += c[j];
    O = (O + t * POWER(K, MOD - 1 + i - N)) % MOD;
  }
  printf("%lld\n", O);
  return 0;
}

### Prompt

Generate 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 long long MOD = 1000000007;
int N, K;
long long O;
int m[100001], pr[100001], v, u[100001], c[100001];
std::vector<int> d[200001];
long long POWER(long long a, long long b) {
  long long r = 1;
  for (; b; b >>= 1) {
    if (b & 1) r = r * a % MOD;
    a = a * a % MOD;
  }
  return r;
}
int main() {
  scanf("%d%d", &N, &K);
  u[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (!m[i]) {
      m[i] = pr[++v] = i;
      u[i] = -1;
    }
    for (int j = 1; j <= v && pr[j] * i <= N && pr[j] <= m[i]; j++) {
      m[pr[j] * i] = pr[j];
      u[pr[j] * i] = pr[j] == m[i] ? 0 : -u[i];
    }
  }
  for (int i = 1; i < N + N; i++)
    for (int j = i; j < N + N; j += i) d[j].push_back(i);
  for (int i = 1; i < N; i++) {
    long long t = 0;
    for (int j = 1; j <= (N - 1) / i; j++)
      if (u[j]) {
        int e = (N - 1) / i / j, r = (N + i) / i / j;
        t += u[j] *
             ((long long)e * e - (long long)(e + e - r) * (e + e - r + 1) / 2);
      }
    std::vector<int> &V = d[N + i];
    int S = V.size();
    for (int j = 0; j < S; j++) c[j] = (N - 1) / V[j] - i / V[j];
    for (int j = S - 1; j >= 0; j--)
      for (int k = j + 1; k < S; k++)
        if (V[k] % V[j] == 0) c[j] -= c[k];
    for (int j = 0; j < S; j++)
      if (V[j] < i) t += c[j];
    O = (O + t * POWER(K, MOD - 1 + i - N)) % MOD;
  }
  printf("%lld\n", O);
  return 0;
}

```