#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 100;
const long long mod = 1e9 + 7;
const long long inf = 1e18;
long long pw(long long a, long long b) {
  long long c = 1;
  while (b) {
    if (b & 1) c = c * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return c;
}
long long n, k, p[maxn], dp[maxn], ans;
vector<long long> g[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  for (long long i = 0; i < maxn; i++) {
    g[i].reserve(10);
  }
  for (long long i = maxn - 1; i > 0; i--) {
    for (long long j = i; j < maxn; j += i) {
      g[j].push_back(i);
    }
  }
  cin >> n >> k;
  p[0] = 1;
  for (long long i = 1; i < maxn; i++) {
    p[i] = p[i - 1] * k % mod;
  }
  for (long long s = 2; s <= n + n - 2; s++) {
    long long l = max(0ll, s - n);
    long long r = min(s - 1, n - 1);
    for (long long i : g[s]) {
      dp[i] = 0;
    }
    for (long long i : g[s]) {
      dp[i] += r / i - l / i;
      ans = (ans + dp[i] * p[max(i, s - n)]) % mod;
      for (long long j : g[i]) {
        if (i != j) dp[j] -= dp[i];
      }
    }
  }
  cout << ans * pw(p[n], mod - 2) % mod;
  return 0;
}

### 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 long long maxn = 2e5 + 100;
const long long mod = 1e9 + 7;
const long long inf = 1e18;
long long pw(long long a, long long b) {
  long long c = 1;
  while (b) {
    if (b & 1) c = c * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return c;
}
long long n, k, p[maxn], dp[maxn], ans;
vector<long long> g[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  for (long long i = 0; i < maxn; i++) {
    g[i].reserve(10);
  }
  for (long long i = maxn - 1; i > 0; i--) {
    for (long long j = i; j < maxn; j += i) {
      g[j].push_back(i);
    }
  }
  cin >> n >> k;
  p[0] = 1;
  for (long long i = 1; i < maxn; i++) {
    p[i] = p[i - 1] * k % mod;
  }
  for (long long s = 2; s <= n + n - 2; s++) {
    long long l = max(0ll, s - n);
    long long r = min(s - 1, n - 1);
    for (long long i : g[s]) {
      dp[i] = 0;
    }
    for (long long i : g[s]) {
      dp[i] += r / i - l / i;
      ans = (ans + dp[i] * p[max(i, s - n)]) % mod;
      for (long long j : g[i]) {
        if (i != j) dp[j] -= dp[i];
      }
    }
  }
  cout << ans * pw(p[n], mod - 2) % mod;
  return 0;
}

```