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;
void read(int &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -f;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  x *= f;
}
void print(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (!x) return;
  print(x / 10), putchar(x % 10 + 48);
}
void write(int x) {
  if (!x)
    putchar('0');
  else
    print(x);
  putchar('\n');
}
const int maxn = 2e5 + 10;
const int inf = 1e9;
const double eps = 1e-8;
const int mod = 1e9 + 7;
int n, k, pri[maxn], tot, mu[maxn], vis[maxn], pw[maxn], cnt[maxn];
void sieve() {
  mu[1] = 1;
  for (int i = 2; i < maxn; i++) {
    if (!vis[i]) pri[++tot] = i, mu[i] = -1;
    for (int j = 1; j <= tot && i * pri[j] < maxn; j++) {
      vis[i * pri[j]] = 1;
      if (i % pri[j] == 0) break;
      mu[i * pri[j]] = -mu[i];
    }
  }
}
int qpow(int a, int x) {
  int res = 1;
  for (; x; x >>= 1, a = 1ll * a * a % mod)
    if (x & 1) res = 1ll * res * a % mod;
  return res;
}
int main() {
  read(n), read(k);
  sieve();
  pw[0] = 1;
  int ans = 0;
  for (int i = 1; i <= n; i++) pw[i] = 1ll * pw[i - 1] * k % mod;
  for (int d = 1; d <= n - 1; d++) {
    int m = (2 * n - 2) / d;
    for (int t = 1; t <= m; t++)
      for (int s = t; s <= m; s += t)
        cnt[s] +=
            mu[t] * (min(s - 1, (n - 1) / d) / t - max(0, (s * d - n) / d) / t);
    for (int s = 1; s <= m; s++)
      ans = (ans + 1ll * pw[max(s * d - n, d)] * cnt[s] % mod) % mod;
    for (int i = 0; i <= m + 2; i++) cnt[i] = 0;
  }
  ans = 1ll * ans * qpow(qpow(k, n), mod - 2) % mod;
  write(ans);
  return 0;
}