#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 1e5 + 9;
long long qpow(long long x, long long y) {
  long long t = 1;
  while (y) {
    if (y & 1) (t *= x) %= mod;
    (x *= x) %= mod;
    y >>= 1;
  }
  return t;
}
int n;
long long k, kk[N];
int f[N], pri[N], tot;
long long phi[N], sph[N];
void init() {
  f[1] = 1, tot = 0, phi[1] = 1;
  for (int i = 2; i <= N - 9; i++) {
    if (!f[i]) pri[++tot] = i, phi[i] = i - 1;
    for (int j = 1; j <= tot && pri[j] * i <= N - 9; j++) {
      f[pri[j] * i] = 1, phi[pri[j] * i] = phi[i] * (pri[j] - 1);
      if (i % pri[j] == 0) {
        phi[pri[j] * i] = phi[i] * pri[j];
        break;
      }
    }
  }
  for (int i = 2; i <= N - 9; i++) sph[i] = (sph[i - 1] + phi[i]) % mod;
  return;
}
int main() {
  init();
  scanf("%d%lld", &n, &k);
  kk[0] = 1;
  for (int i = 1; i <= n; i++) kk[i] = kk[i - 1] * k % mod;
  long long ans = 0;
  for (int i = 1; i < (n - 1); i++) {
    (ans += kk[i] * (n - i - 1) % mod) %= mod;
  }
  for (int p = 1; p <= n - 1; p++) {
    (ans += sph[(n - 1) / p + 1] * kk[p] % mod) %= mod;
  }
  for (int p = 1; p <= n - 1; p++) {
    if (n % p == 0) continue;
    (ans +=
     (mod - phi[(n - 1) / p + 1] * kk[((n - 1) / p + 1) * p - n] % mod)) %= mod;
  }
  printf("%lld\n", ans * qpow(kk[n], mod - 2) % mod);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 mod = 1e9 + 7;
const int N = 1e5 + 9;
long long qpow(long long x, long long y) {
  long long t = 1;
  while (y) {
    if (y & 1) (t *= x) %= mod;
    (x *= x) %= mod;
    y >>= 1;
  }
  return t;
}
int n;
long long k, kk[N];
int f[N], pri[N], tot;
long long phi[N], sph[N];
void init() {
  f[1] = 1, tot = 0, phi[1] = 1;
  for (int i = 2; i <= N - 9; i++) {
    if (!f[i]) pri[++tot] = i, phi[i] = i - 1;
    for (int j = 1; j <= tot && pri[j] * i <= N - 9; j++) {
      f[pri[j] * i] = 1, phi[pri[j] * i] = phi[i] * (pri[j] - 1);
      if (i % pri[j] == 0) {
        phi[pri[j] * i] = phi[i] * pri[j];
        break;
      }
    }
  }
  for (int i = 2; i <= N - 9; i++) sph[i] = (sph[i - 1] + phi[i]) % mod;
  return;
}
int main() {
  init();
  scanf("%d%lld", &n, &k);
  kk[0] = 1;
  for (int i = 1; i <= n; i++) kk[i] = kk[i - 1] * k % mod;
  long long ans = 0;
  for (int i = 1; i < (n - 1); i++) {
    (ans += kk[i] * (n - i - 1) % mod) %= mod;
  }
  for (int p = 1; p <= n - 1; p++) {
    (ans += sph[(n - 1) / p + 1] * kk[p] % mod) %= mod;
  }
  for (int p = 1; p <= n - 1; p++) {
    if (n % p == 0) continue;
    (ans +=
     (mod - phi[(n - 1) / p + 1] * kk[((n - 1) / p + 1) * p - n] % mod)) %= mod;
  }
  printf("%lld\n", ans * qpow(kk[n], mod - 2) % mod);
  return 0;
}

```