#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100000;
const int MOD = int(1E9) + 7;
int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }
int pow_mod(int b, int p) {
  int ret = 1;
  while (p) {
    if (p & 1) ret = 1LL * ret * b % MOD;
    b = 1LL * b * b % MOD;
    p >>= 1;
  }
  return ret;
}
int n, m, k, ans;
void update(int d, int s, int p) {
  ans = (ans + 1LL * p * pow_mod(k, MOD - 1 + max(s - n, d) - n) % MOD) % MOD;
}
vector<int> dv[2 * MAXN + 5];
void init() {
  for (int i = 1; i <= m; i++)
    for (int j = i; j <= m; j += i) dv[j].push_back(i);
}
int f[2 * MAXN + 5];
int main() {
  scanf("%d%d", &n, &k), m = 2 * (n - 1);
  init();
  for (int s = 1; s <= m; s++) {
    int l = max(1, s - (n - 1)), r = min(n - 1, s - 1);
    for (int j = dv[s].size() - 1; j >= 0; j--) {
      int d = dv[s][j], p = s / d;
      f[d] = r / d - (l - 1) / d;
      for (int k = 1; k < dv[p].size(); k++)
        f[d] = (f[d] + MOD - f[d * dv[p][k]]) % MOD;
      update(d, s, f[d]);
    }
  }
  printf("%d\n", ans);
}

### 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>
using namespace std;
const int MAXN = 100000;
const int MOD = int(1E9) + 7;
int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }
int pow_mod(int b, int p) {
  int ret = 1;
  while (p) {
    if (p & 1) ret = 1LL * ret * b % MOD;
    b = 1LL * b * b % MOD;
    p >>= 1;
  }
  return ret;
}
int n, m, k, ans;
void update(int d, int s, int p) {
  ans = (ans + 1LL * p * pow_mod(k, MOD - 1 + max(s - n, d) - n) % MOD) % MOD;
}
vector<int> dv[2 * MAXN + 5];
void init() {
  for (int i = 1; i <= m; i++)
    for (int j = i; j <= m; j += i) dv[j].push_back(i);
}
int f[2 * MAXN + 5];
int main() {
  scanf("%d%d", &n, &k), m = 2 * (n - 1);
  init();
  for (int s = 1; s <= m; s++) {
    int l = max(1, s - (n - 1)), r = min(n - 1, s - 1);
    for (int j = dv[s].size() - 1; j >= 0; j--) {
      int d = dv[s][j], p = s / d;
      f[d] = r / d - (l - 1) / d;
      for (int k = 1; k < dv[p].size(); k++)
        f[d] = (f[d] + MOD - f[d * dv[p][k]]) % MOD;
      update(d, s, f[d]);
    }
  }
  printf("%d\n", ans);
}

```