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 = 2e5 + 5;
const int P = 1e9 + 7;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
int power(int x, int y) {
  if (y == 0) return 1;
  int tmp = power(x, y / 2);
  if (y % 2 == 0)
    return 1ll * tmp * tmp % P;
  else
    return 1ll * tmp * tmp % P * x % P;
}
void update(int &x, int y) {
  x += y;
  if (x >= P) x -= P;
}
int miu[MAXN], powk[MAXN];
vector<int> factors[MAXN];
int tot, prime[MAXN], f[MAXN];
void sieve(int n) {
  miu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (f[i] == 0) prime[++tot] = f[i] = i, miu[i] = P - 1;
    for (int j = 1; j <= tot && prime[j] <= f[i]; j++) {
      int tmp = prime[j] * i;
      if (tmp > n) break;
      if (prime[j] == f[i])
        miu[tmp] = 0;
      else
        miu[tmp] = (P - miu[i]) % P;
      f[tmp] = prime[j];
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j += i) factors[j].push_back(i);
}
int calc(int lim, int sum) {
  int l = 1, r = lim;
  chkmax(l, sum - lim);
  chkmin(r, sum - 1);
  if (l > r) return 0;
  int ans = 0;
  for (auto x : factors[sum])
    update(ans, 1ll * miu[x] * (P - (l - 1) / x + r / x) % P);
  return ans;
}
int main() {
  int n, k, ans = 0;
  read(n), read(k);
  sieve(2 * n), powk[0] = 1;
  for (int i = 1; i <= n; i++) powk[i] = 1ll * powk[i - 1] * k % P;
  for (int s = 2; s <= 2 * n - 2; s++)
    for (auto g : factors[s]) {
      int lim = (n - 1) / g, sum = s / g;
      if (lim <= 0 || sum <= 1) continue;
      update(ans, 1ll * calc(lim, sum) * powk[max(g, s - n)] % P);
    }
  writeln(1ll * ans * power(powk[n], P - 2) % P);
  return 0;
}