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 int N = 1e5 + 5, mo = 1e9 + 7;
inline int poww(int x, long long y) {
  int ans = 1;
  for (; y; y >>= 1, x = 1ll * x * x % mo)
    if (y & 1) ans = 1ll * ans * x % mo;
  return ans;
}
int n, k, i, j, l, x, ans;
int p[N / 5], xb, phi[N], ps[N];
bool b[N];
int g[N];
int gfa(int x) { return g[x] == x ? x : g[x] = gfa(g[x]); }
int main() {
  for (i = 2; i < N; ++i) {
    if (!b[i]) p[++xb] = i, phi[i] = i - 1;
    for (j = 1; j <= xb; ++j) {
      x = i * p[j];
      if (x >= N) break;
      b[x] = 1;
      if (i % p[j] == 0) {
        phi[x] = phi[i] * p[j];
        break;
      }
      phi[x] = phi[i] * (p[j] - 1);
    }
  }
  for (i = phi[1] = 1; i < N; ++i) ps[i] = (ps[i - 1] + phi[i]) % mo;
  scanf("%d%d", &n, &k);
  for (i = 1; i < n; ++i) {
    int d = (n - 1) / i;
    ans = (ans + 1ll * poww(k, i) * (ps[d + 1] - 1)) % mo;
    ans = (ans + mo - 1ll * poww(k, (d + 1) * i - n) * phi[d + 1] % mo) % mo;
  }
  for (i = n; i <= n * 2 - 2; ++i)
    ans = (ans + 1ll * poww(k, i - n) * (n * 2 - 1 - i)) % mo;
  ans = 1ll * ans * poww(poww(k, n), mo - 2) % mo;
  printf("%d\n", ans);
  return 0;
}