#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long MOD = (long long)1e9 + 7;
long long add(long long x, long long y) {
  x += y;
  if (x >= MOD) return x - MOD;
  return x;
}
long long mult(long long x, long long y) {
  return ((x % MOD) * (y % MOD)) % MOD;
}
long long bin_pow(long long x, long long p) {
  if (p == 0) return 1;
  if (p & 1) return mult(x, bin_pow(x, p - 1));
  return bin_pow(mult(x, x), p / 2);
}
long long rev(long long x) { return bin_pow(x, MOD - 2); }
const int N = 200200;
int n;
long long k;
long long ans[N];
vector<int> d[N];
long long c[N];
long long dp[N];
long long mu[N];
long long prefMu[N];
void precalc() {
  for (int x = 1; x < N; x++)
    for (int y = x; y < N; y += x) d[y].push_back(x);
}
int main() {
  precalc();
  scanf("%d%lld", &n, &k);
  for (int s = n + 1; s < 2 * n - 1; s++) {
    int l = s - n, r = n - 1;
    for (int x : d[s]) {
      c[x] = (r / x) - (l / x);
    }
    for (int i = (int)d[s].size() - 1; i >= 0; i--) {
      int x = d[s][i];
      for (int y : d[x])
        if (y != x) c[y] -= c[x];
    }
    for (int g : d[s]) {
      if (s - g >= n) ans[s - n] += c[g];
    }
  }
  for (int m = 2; m <= n; m++) {
    dp[m] = (long long)m * (m - 1) / 2;
    long long b = (long long)sqrt(m);
    while (b * b < m) b++;
    for (int i = 2; i <= b; i++) dp[m] -= dp[m / i];
    for (int i = b; i >= 1; i--) {
      long long z = m / i;
      if (z > b) {
        dp[m] -= (z - b) * dp[i];
        b = z;
      }
    }
  }
  for (int g = 1; g < n; g++) {
    ans[g] += dp[(n - 1) / g + 1];
  }
  long long res = 0;
  for (int i = 1; i < n; i++) {
    ans[i] %= MOD;
    res = add(res, mult(ans[i], bin_pow(k, i)));
  }
  res = mult(res, rev(bin_pow(k, n)));
  printf("%lld\n", res);
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long MOD = (long long)1e9 + 7;
long long add(long long x, long long y) {
  x += y;
  if (x >= MOD) return x - MOD;
  return x;
}
long long mult(long long x, long long y) {
  return ((x % MOD) * (y % MOD)) % MOD;
}
long long bin_pow(long long x, long long p) {
  if (p == 0) return 1;
  if (p & 1) return mult(x, bin_pow(x, p - 1));
  return bin_pow(mult(x, x), p / 2);
}
long long rev(long long x) { return bin_pow(x, MOD - 2); }
const int N = 200200;
int n;
long long k;
long long ans[N];
vector<int> d[N];
long long c[N];
long long dp[N];
long long mu[N];
long long prefMu[N];
void precalc() {
  for (int x = 1; x < N; x++)
    for (int y = x; y < N; y += x) d[y].push_back(x);
}
int main() {
  precalc();
  scanf("%d%lld", &n, &k);
  for (int s = n + 1; s < 2 * n - 1; s++) {
    int l = s - n, r = n - 1;
    for (int x : d[s]) {
      c[x] = (r / x) - (l / x);
    }
    for (int i = (int)d[s].size() - 1; i >= 0; i--) {
      int x = d[s][i];
      for (int y : d[x])
        if (y != x) c[y] -= c[x];
    }
    for (int g : d[s]) {
      if (s - g >= n) ans[s - n] += c[g];
    }
  }
  for (int m = 2; m <= n; m++) {
    dp[m] = (long long)m * (m - 1) / 2;
    long long b = (long long)sqrt(m);
    while (b * b < m) b++;
    for (int i = 2; i <= b; i++) dp[m] -= dp[m / i];
    for (int i = b; i >= 1; i--) {
      long long z = m / i;
      if (z > b) {
        dp[m] -= (z - b) * dp[i];
        b = z;
      }
    }
  }
  for (int g = 1; g < n; g++) {
    ans[g] += dp[(n - 1) / g + 1];
  }
  long long res = 0;
  for (int i = 1; i < n; i++) {
    ans[i] %= MOD;
    res = add(res, mult(ans[i], bin_pow(k, i)));
  }
  res = mult(res, rev(bin_pow(k, n)));
  printf("%lld\n", res);
  return 0;
}

```