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 maxn = 400010, mod = 1e9 + 7;
long long n, k, ans, miu[maxn], pri[maxn], cnt, v[maxn], pk[maxn];
long long Power(long long x, long long p) {
  long long res = 1;
  for (; p; p >>= 1, x = x * x % mod)
    if (p & 1) res = res * x % mod;
  return res;
}
int main() {
  scanf("%lld%lld", &n, &k);
  k = Power(k, mod - 2);
  miu[1] = 1;
  v[1] = 1;
  for (int i(2), i_ORZ(n); i <= i_ORZ; i++) {
    if (!v[i]) pri[++cnt] = i, miu[i] = -1;
    for (int j = 1; j <= cnt && i * pri[j] <= n; j++) {
      v[i * pri[j]] = 1;
      if (i % pri[j])
        miu[i * pri[j]] = -miu[i];
      else
        miu[i * pri[j]] = 0;
    }
  }
  for (int i(0), i_ORZ(n * 2); i <= i_ORZ; i++) pk[i] = Power(k, i);
  for (int d(1), d_ORZ(n - 1); d <= d_ORZ; d++)
    for (int t(1), t_ORZ((n * 2 - 2) / d); t <= t_ORZ; t++)
      for (long long s = t * d; s <= n * 2 - 2; s += t * d) {
        if (s == d) continue;
        long long tmp =
            miu[t] * (min(n - 1, s - 1) / d / t - max(0ll, s - n) / d / t);
        ans = (ans + tmp * pk[k, n - max(1ll * d, s - n)] % mod + mod) % mod;
      }
  printf("%lld\n", ans);
  return 0;
}