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>
const int N = 100005, mod = 1000000007;
int n, k, md[N], phi[N], pr[N], pt, ans, cnt[N], sum[N];
void up(int &x, int y) { x += y - mod, x += x >> 31 & mod; }
int pow(int x, int y, int ans = 1) {
  for (; y; y >>= 1, x = (long long)x * x % mod)
    if (y & 1) ans = (long long)ans * x % mod;
  return ans;
}
void sieve(int n) {
  for (int i = 2; i <= n; ++i) {
    if (!md[i]) md[i] = i, pr[++pt] = i, phi[i] = i - 1;
    for (int j = 1, t; j <= pt && (t = i * pr[j]) <= n; ++j) {
      md[t] = pr[j];
      if (pr[j] == md[i]) {
        phi[t] = phi[i] * pr[j];
        break;
      } else
        phi[t] = phi[i] * phi[pr[j]];
    }
  }
  for (int i = 2; i <= n; ++i) up(sum[i] = sum[i - 1], phi[i]);
}
int main() {
  std::ios::sync_with_stdio(0), std::cin.tie(0);
  std::cin >> n >> k, sieve(n);
  for (int i = 1; i < n; ++i) cnt[i] = n - i - 1;
  for (int i = 1; i < n; ++i) {
    int d = (n + i - 1) / i;
    up(cnt[i], sum[d]);
    up(cnt[d * i - n], mod - phi[d]);
  }
  for (int i = 1, t = 1; i < n; ++i) {
    t = (long long)t * k % mod;
    ans = (ans + (long long)t * cnt[i]) % mod;
  }
  std::cout << pow(k, mod - 1 - n, ans) << '\n';
  return 0;
}