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>
int n, m, k, ik, f[200005], ans;
std ::vector<int> d[200005];
int qpow(int a, int b) {
  int s = 1;
  for (; b; b >>= 1, a = 1ll * a * a % 1000000007)
    if (b & 1) s = 1ll * s * a % 1000000007;
  return s;
}
void inc(int &a, int b) { (a += b) >= 1000000007 ? a -= 1000000007 : 0; }
void dec(int &a, int b) { (a -= b) < 0 ? a += 1000000007 : 0; }
int main() {
  scanf("%d%d", &n, &k), ik = qpow(k, 1000000007 - 2), m = (n - 1) << 1;
  for (register int i = 1; i <= m; ++i)
    for (register int j = i; j <= m; j += i) d[j].push_back(i);
  for (register int s = 1; s <= m; ++s) {
    int l = std ::max(s - n + 1, 1), r = std ::min(s - 1, n - 1);
    for (register int i = d[s].size() - 1; ~i; --i) {
      int g = d[s][i];
      f[g] = r / g - (l - 1) / g;
      for (int p : d[s / g])
        if (p > 1) dec(f[g], f[g * p]);
      if (g <= n)
        inc(ans, 1ll * f[g] * qpow(ik, n - std ::max(s - n, g)) % 1000000007);
    }
  }
  printf("%d\n", ans);
}