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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2, md) % md
                      : power(a * a % md, b / 2, md) % md));
}
const int xn = 2e5 + 10;
const int xm = -20 + 10;
const int sq = 320;
const int inf = 1e9 + 10;
const long long INF = 1e18 + 10;
const long long mod = 1e9 + 7;
const int base = 257;
int n, k, pw[xn], ans, p[xn];
vector<pair<int, int> > dp[xn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  for (int i = n + n; i >= 1; --i) {
    for (int j = 1; j * i < n + n; ++j) {
      p[i * j] = j - 1;
      dp[i].push_back({i * j, max(0, min(j - 1, (n - 1) / i) -
                                         max(1, j - (n - 1) / i) + 1)});
    }
    for (int j = i + i; j < n + n; j += i)
      for (pair<int, int> k : dp[j]) dp[i][p[k.first]].second -= k.second;
  }
  pw[0] = 1;
  for (int i = 1; i < xn; ++i) pw[i] = 1LL * pw[i - 1] * k % mod;
  for (int i = 1; i < n; ++i)
    for (pair<int, int> j : dp[i])
      ans = (ans + 1LL * pw[max(i, j.first - n)] * j.second % mod) % mod;
  cout << 1LL * ans * power(pw[n], mod - 2, mod) % mod << '\n';
  return 0;
}