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 = 1e5 + 5, M = N << 1, rqy = 1e9 + 7;
int n, k, tot, pri[N], miu[N], pk[M], ik, ipk[M], ans;
bool mark[N];
std::vector<int> k1[N], k2[N];
int qpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % rqy;
    a = 1ll * a * a % rqy;
    b >>= 1;
  }
  return res;
}
void sieve() {
  mark[0] = mark[miu[1] = 1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!mark[i]) miu[pri[++tot] = i] = -1;
    for (int j = 1; j <= tot; j++) {
      if (1ll * i * pri[j] > n) break;
      mark[i * pri[j]] = 1;
      if (i % pri[j] == 0) break;
      miu[i * pri[j]] = -miu[i];
    }
  }
}
int czx(int d, int p, int m) {
  int x = (n - 1) / p, y = m / (p / d);
  int res = (1ll * k1[p][x] * k1[p][x] - k2[p][y] + rqy) % rqy;
  return (1ll * res * ipk[n] + 1ll * y * (y - 1) / 2 % rqy * pk[d]) % rqy;
}
int main() {
  std::cin >> n >> k;
  sieve();
  pk[0] = ipk[0] = 1;
  ik = qpow(k, rqy - 2);
  for (int i = 1; i <= (n << 1); i++)
    pk[i] = 1ll * pk[i - 1] * k % rqy, ipk[i] = 1ll * ipk[i - 1] * ik % rqy;
  for (int d = 1; d <= n; d++) {
    k1[d].push_back(0);
    k2[d].push_back(0);
    for (int i = 1; i <= n / d + 1; i++)
      k1[d].push_back((k1[d][i - 1] + pk[d * i]) % rqy),
          k2[d].push_back((1ll * (i - 1) * pk[d * i] + k2[d][i - 1]) % rqy);
  }
  for (int d = 1; d < n; d++) {
    int m = (n + d - 1) / d;
    for (int i = d; i < n; i += d)
      ans = (1ll * czx(d, i, m) * miu[i / d] + ans + rqy) % rqy;
  }
  return std::cout << 1ll * ans * ipk[n] % rqy << std::endl, 0;
}