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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && (c != '-')) c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
const int N = 1e5 + 10, mod = 1e9 + 7;
int n, k, p[2 * N];
inline int power(int a, int b) {
  int ret = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ret = 1ll * ret * a % mod;
  return ret;
}
int tot, pri[N], mu[N];
bool vis[N];
inline void init(int n) {
  mu[1] = 1;
  for (register int i = (2); i <= (n); i++) {
    if (!vis[i]) pri[++tot] = i, mu[i] = mod - 1;
    for (int j = 1; j <= tot && i * pri[j] <= n; ++j) {
      vis[i * pri[j]] = 1;
      if (i % pri[j] == 0) break;
      mu[i * pri[j]] = mod - mu[i];
    }
  }
}
int inv[2 * N];
inline int calc(int y, int x, int a) {
  if (y < 0) return 0;
  return 1ll * p[a] * (1ll * (p[x * (y + 1)] + mod - 1) * inv[x] % mod) % mod;
}
int main() {
  n = read(), k = read();
  if (k == 1) {
    printf("%d\n", 1ll * (n - 1) * (n - 1) % mod);
    return 0;
  }
  p[0] = 1;
  for (register int i = (1); i <= (2 * n); i++)
    p[i] = 1ll * p[i - 1] * k % mod, inv[i] = power(p[i] - 1, mod - 2);
  init(n);
  int ans = 0;
  for (register int d = (1); d < (n); d++)
    for (register int _d = (1); _d <= ((n - 1) / d); _d++)
      if (mu[_d])
        for (register int i = (1); i <= ((n - 1) / d / _d); i++) {
          int k = (n + d) / (d * _d) - i + 1;
          ans = (ans +
                 1ll * mu[_d] *
                     calc((n - 1) / d / _d - k, d * _d, (i + k) * d * _d - n)) %
                mod;
          ans = (ans + 1ll * mu[_d] * max(0, min((n - 1) / d / _d, k - 1)) %
                           mod * p[d]) %
                mod;
        }
  printf("%lld\n", 1ll * ans * power(p[n], mod - 2) % mod);
}