#include <bits/stdc++.h>
using ll = long long;
const int MN = 1e5 + 100;
const int MP = 1e4 + 100;
const int MOD = 1e9 + 7;
ll pw(ll b, int p) {
  ll r = 1LL;
  for (; p; p >>= 1, b = b * b % MOD)
    if (p & 1) r = r * b % MOD;
  return r;
}
int N, K;
int p[MP], q[MN];
int t[MN], l[MN], T[MN];
void pcpp(void) {
  for (int i = 2; i <= N; ++i) {
    if (!q[i]) t[i] = i - 1, p[l[i] = ++p[0]] = i;
    for (int j = 1, k; j <= p[0] && (k = i * p[j]) <= N; ++j) {
      assert(!q[k]), q[k] = 1, l[k] = j;
      if (l[i] == j) {
        t[k] = t[i] * p[j];
        break;
      }
      t[k] = t[i] * (p[j] - 1);
    }
  }
}
ll v[MN], f, z[MN];
int main(void) {
  scanf("%d%d", &N, &K);
  pcpp();
  z[0] = 1LL, z[1] = pw(K, MOD - 2);
  for (int i = 1; i < N; ++i) z[i + 1] = z[i] * z[1] % MOD;
  for (int i = 0; i <= N; ++i) T[i + 1] = (t[i] + T[i]) % MOD;
  for (int i = 1; i < N; ++i) v[i] = N - i - 1;
  for (int g = 1; g < N; ++g) {
    int k = (N + g - 1) / g;
    v[g] += T[k + 1];
    if (k * g > N) v[k * g - N] -= t[k];
  }
  for (int i = 1; i < N; ++i) f += v[i] * z[N - i] % MOD;
  printf("%lld\n", f % MOD);
  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>
using ll = long long;
const int MN = 1e5 + 100;
const int MP = 1e4 + 100;
const int MOD = 1e9 + 7;
ll pw(ll b, int p) {
  ll r = 1LL;
  for (; p; p >>= 1, b = b * b % MOD)
    if (p & 1) r = r * b % MOD;
  return r;
}
int N, K;
int p[MP], q[MN];
int t[MN], l[MN], T[MN];
void pcpp(void) {
  for (int i = 2; i <= N; ++i) {
    if (!q[i]) t[i] = i - 1, p[l[i] = ++p[0]] = i;
    for (int j = 1, k; j <= p[0] && (k = i * p[j]) <= N; ++j) {
      assert(!q[k]), q[k] = 1, l[k] = j;
      if (l[i] == j) {
        t[k] = t[i] * p[j];
        break;
      }
      t[k] = t[i] * (p[j] - 1);
    }
  }
}
ll v[MN], f, z[MN];
int main(void) {
  scanf("%d%d", &N, &K);
  pcpp();
  z[0] = 1LL, z[1] = pw(K, MOD - 2);
  for (int i = 1; i < N; ++i) z[i + 1] = z[i] * z[1] % MOD;
  for (int i = 0; i <= N; ++i) T[i + 1] = (t[i] + T[i]) % MOD;
  for (int i = 1; i < N; ++i) v[i] = N - i - 1;
  for (int g = 1; g < N; ++g) {
    int k = (N + g - 1) / g;
    v[g] += T[k + 1];
    if (k * g > N) v[k * g - N] -= t[k];
  }
  for (int i = 1; i < N; ++i) f += v[i] * z[N - i] % MOD;
  printf("%lld\n", f % MOD);
  return 0;
}

```