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;
const long long M = 1000000007;
const int maxn = 211111;
long long n, k, p[maxn], ans, cnt[maxn], phi[maxn], mu[maxn], f[maxn];
vector<int> h[maxn];
void cc(int x, int y) {
  for (auto v : h[x]) {
    int r = y / v;
    for (auto vv : h[v]) cnt[vv] += mu[v / vv] * r;
  }
}
long long pow_(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % M;
    x = x * x % M;
    y >>= 1;
  }
  return ret;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < maxn; i++) phi[i] = i;
  mu[1] = 1;
  for (int i = 1; i < maxn; i++)
    for (int j = 2; j * i < maxn; j++) phi[i * j] -= phi[i], mu[i * j] -= mu[i];
  for (int i = 1; i < maxn; i++)
    for (int j = 1; j * i < maxn; j++) h[i * j].push_back(i);
  for (int i = 1; i < n; i++) cc(i, n - i);
  for (int i = n + 1; i <= n * 2 - 2; i++) {
    int l, r;
    l = i - n;
    r = n - 1;
    for (auto v : h[i]) {
      int t = r / v - l / v;
      for (auto vv : h[v]) f[vv] += mu[v / vv] * t;
    }
    for (auto v : h[i]) {
      cnt[max(v, i - (int)n)] += f[v];
      f[v] = 0;
    }
  }
  for (int i = 0; i < maxn; i++) cnt[i] %= M;
  for (int i = 0; i < maxn; i++) (ans += pow_(k, i) * cnt[i]) %= M;
  long long inv = pow_(k, M - 2);
  ans = ans * pow_(inv, n) % M;
  cout << ans << endl;
  return 0;
}