#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
const long long MOD = 1e9 + 7;
int prime[MAXN];
long long phis[MAXN], powk[MAXN];
long long mpow(long long b, long long e) {
  long long res = 1;
  for (long long k = 1; k <= e; k *= 2) {
    if (k & e) res = (res * b) % MOD;
    b = (b * b) % MOD;
  }
  return res;
}
long long inv(long long b) { return mpow(b, MOD - 2); }
vector<int> prod(vector<int> a, int x) {
  vector<int> res = a;
  for (auto y : a) res.push_back(x * y);
  return res;
}
long long f(int y, int n) {
  if (n == 0) return 0;
  int x = y;
  vector<int> facs = {1};
  while (x > 1) {
    int p = prime[x];
    facs = prod(facs, -p);
    while (x % p == 0) x /= p;
  }
  int res = 0;
  for (auto f : facs) {
    int k = n / abs(f);
    if (f > 0)
      res += k;
    else
      res -= k;
  }
  return res;
}
void sieve() {
  for (int p = 2; p < MAXN; p++) {
    if (prime[p]) continue;
    for (int q = p; q < MAXN; q += p) prime[q] = p;
  }
}
long long n, k;
long long calc(long long s, long long d) {
  if (s <= n - 1)
    return f(s / d, s / d - 1) * powk[max(s - n, d)];
  else
    return (f(s / d, (n - 1) / d) - f(s / d, (s - n) / d)) *
           powk[max(s - n, d)];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  sieve();
  long long ans = 0;
  cin >> n >> k;
  powk[0] = 1;
  for (int i = 1; i < MAXN; i++) powk[i] = (k * powk[i - 1]) % MOD;
  for (long long s = 2; s <= 2 * n - 2; s++) {
    for (long long d = 1; d * d <= s; d++) {
      if (s % d) continue;
      ans = (ans + calc(s, d)) % MOD;
      if (s / d > d) {
        long long d1 = s / d;
        ans = (ans + calc(s, d1)) % MOD;
      }
    }
  }
  ans = (ans * inv(powk[n])) % MOD;
  cout << ans << '\n';
}

### Prompt

In cpp, your task is to solve 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 = 2e5 + 5;
const long long MOD = 1e9 + 7;
int prime[MAXN];
long long phis[MAXN], powk[MAXN];
long long mpow(long long b, long long e) {
  long long res = 1;
  for (long long k = 1; k <= e; k *= 2) {
    if (k & e) res = (res * b) % MOD;
    b = (b * b) % MOD;
  }
  return res;
}
long long inv(long long b) { return mpow(b, MOD - 2); }
vector<int> prod(vector<int> a, int x) {
  vector<int> res = a;
  for (auto y : a) res.push_back(x * y);
  return res;
}
long long f(int y, int n) {
  if (n == 0) return 0;
  int x = y;
  vector<int> facs = {1};
  while (x > 1) {
    int p = prime[x];
    facs = prod(facs, -p);
    while (x % p == 0) x /= p;
  }
  int res = 0;
  for (auto f : facs) {
    int k = n / abs(f);
    if (f > 0)
      res += k;
    else
      res -= k;
  }
  return res;
}
void sieve() {
  for (int p = 2; p < MAXN; p++) {
    if (prime[p]) continue;
    for (int q = p; q < MAXN; q += p) prime[q] = p;
  }
}
long long n, k;
long long calc(long long s, long long d) {
  if (s <= n - 1)
    return f(s / d, s / d - 1) * powk[max(s - n, d)];
  else
    return (f(s / d, (n - 1) / d) - f(s / d, (s - n) / d)) *
           powk[max(s - n, d)];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  sieve();
  long long ans = 0;
  cin >> n >> k;
  powk[0] = 1;
  for (int i = 1; i < MAXN; i++) powk[i] = (k * powk[i - 1]) % MOD;
  for (long long s = 2; s <= 2 * n - 2; s++) {
    for (long long d = 1; d * d <= s; d++) {
      if (s % d) continue;
      ans = (ans + calc(s, d)) % MOD;
      if (s / d > d) {
        long long d1 = s / d;
        ans = (ans + calc(s, d1)) % MOD;
      }
    }
  }
  ans = (ans * inv(powk[n])) % MOD;
  cout << ans << '\n';
}

```