#include <bits/stdc++.h>
using namespace std;
const long long _ = 1e5 + 7, MOD = 1e9 + 7;
long long N, K, ans = 0, mu[_], prm[_], cnt;
bool nprm[_];
vector<long long> ys[_];
long long add(long long a, long long b) {
  return a + b >= MOD ? a + b - MOD : a + b;
}
long long poww(long long a, long long b) {
  long long times = 1;
  while (b) {
    if (b & 1) times = times * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return times;
}
void init() {
  mu[1] = 1;
  for (long long i = 2; i <= N; ++i) {
    if (!nprm[i]) mu[prm[++cnt] = i] = -1;
    for (long long j = 1; prm[j] * i <= N; ++j) {
      nprm[prm[j] * i] = 1;
      if (i % prm[j] == 0) break;
      mu[prm[j] * i] = -mu[i];
    }
  }
}
signed main() {
  cin >> N >> K;
  init();
  for (long long i = 1; i < N; ++i)
    for (long long j = i; j < N; j += i) ys[j].push_back(i);
  for (long long T = 1; T < N; ++T) {
    static long long sum[_ << 1];
    long long R = (N - 1) / T, sum1 = 0, sum2 = 0;
    sum[2 * R + 1] = 0;
    for (long long j = 2; j <= 2 * R; ++j)
      sum[j] = (j <= R + 1 ? (j - 1) : (2 * R - j + 1)) * poww(K, T * j) % MOD;
    for (long long j = 2 * R - 1; j >= 2; --j) sum[j] = add(sum[j], sum[j + 1]);
    for (auto t : ys[T])
      if (mu[T / t]) {
        long long rge = min(2 * R, (N - 1 + t) / T),
                  area = rge <= R + 1 ? ((rge - 1) * ((rge - 1) + 1) / 2 % MOD)
                                      : R * R - ((2 * R - rge) *
                                                 ((2 * R - rge) + 1) / 2 % MOD);
        sum1 = (sum1 + poww(K, t) * mu[T / t] * area % MOD + MOD) % MOD;
        sum2 = (sum2 + mu[T / t] * sum[rge + 1] + MOD) % MOD;
      }
    ans = (ans + sum1 * poww(poww(K, MOD - 2), N) +
           sum2 * poww(poww(K, MOD - 2), 2 * N)) %
          MOD;
  }
  cout << ans;
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
const long long _ = 1e5 + 7, MOD = 1e9 + 7;
long long N, K, ans = 0, mu[_], prm[_], cnt;
bool nprm[_];
vector<long long> ys[_];
long long add(long long a, long long b) {
  return a + b >= MOD ? a + b - MOD : a + b;
}
long long poww(long long a, long long b) {
  long long times = 1;
  while (b) {
    if (b & 1) times = times * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return times;
}
void init() {
  mu[1] = 1;
  for (long long i = 2; i <= N; ++i) {
    if (!nprm[i]) mu[prm[++cnt] = i] = -1;
    for (long long j = 1; prm[j] * i <= N; ++j) {
      nprm[prm[j] * i] = 1;
      if (i % prm[j] == 0) break;
      mu[prm[j] * i] = -mu[i];
    }
  }
}
signed main() {
  cin >> N >> K;
  init();
  for (long long i = 1; i < N; ++i)
    for (long long j = i; j < N; j += i) ys[j].push_back(i);
  for (long long T = 1; T < N; ++T) {
    static long long sum[_ << 1];
    long long R = (N - 1) / T, sum1 = 0, sum2 = 0;
    sum[2 * R + 1] = 0;
    for (long long j = 2; j <= 2 * R; ++j)
      sum[j] = (j <= R + 1 ? (j - 1) : (2 * R - j + 1)) * poww(K, T * j) % MOD;
    for (long long j = 2 * R - 1; j >= 2; --j) sum[j] = add(sum[j], sum[j + 1]);
    for (auto t : ys[T])
      if (mu[T / t]) {
        long long rge = min(2 * R, (N - 1 + t) / T),
                  area = rge <= R + 1 ? ((rge - 1) * ((rge - 1) + 1) / 2 % MOD)
                                      : R * R - ((2 * R - rge) *
                                                 ((2 * R - rge) + 1) / 2 % MOD);
        sum1 = (sum1 + poww(K, t) * mu[T / t] * area % MOD + MOD) % MOD;
        sum2 = (sum2 + mu[T / t] * sum[rge + 1] + MOD) % MOD;
      }
    ans = (ans + sum1 * poww(poww(K, MOD - 2), N) +
           sum2 * poww(poww(K, MOD - 2), 2 * N)) %
          MOD;
  }
  cout << ans;
  return 0;
}

```