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 ll = long long;
const int MOD = 1e9 + 7;
const ll BMOD = static_cast<ll>(MOD) * MOD;
const int MN = 1e5 + 100;
ll pw(ll b, int p) {
  ll r = 1LL;
  for (; p; p >>= 1, b = b * b % MOD)
    if (p & 1) r = r * b % MOD;
  return r;
}
int N, K;
int T[MN * 2];
std::vector<int> d[MN * 2];
ll f;
ll p[MN];
const int MA = 15;
int a[MA], A;
int v[1 << MA], u[1 << MA];
int count(int v, int l, int r) {
  if (l < 0) l = 0;
  if (r >= v) r = v - 1;
  if (l >= r) return 0;
  int f = r - l;
  A = 0;
  for (int x : d[v])
    if (d[x].empty()) a[A++] = x;
  for (int i = 1, x, y; i < 1 << A; i++) {
    x = 31 - __builtin_clz(i);
    y = ::v[i] = ::v[i ^ 1 << x] * a[x];
    f += u[i] * (r / y - l / y);
  }
  return f;
}
int main(void) {
  scanf("%d%d", &N, &K);
  for (int i = 2; i <= N; i++) {
    for (int j = i * 2; j <= 2 * N; j += i) d[j].push_back(i);
    if (d[i].empty()) {
      T[i] = i - 1;
      for (int j = i * 2; j <= 2 * N; j += i)
        T[j] = T[j] ? T[j] / i * (i - 1) : j / i * (i - 1);
    }
  }
  p[0] = 1LL;
  p[1] = pw(K, MOD - 2);
  for (int i = 1; i < N; i++) p[i + 1] = p[i] * p[1] % MOD;
  v[0] = u[0] = 1;
  for (int i = 1; i < 1 << MA; i++) u[i] = u[i & i - 1] * -1;
  for (int s = 2; s <= N - 1 << 1; s++) {
    for (int x : d[s]) {
      f += p[N - std::max(s - N, x)] * count(s / x, (s - N) / x, (N - 1) / x);
      if (f >= BMOD) f -= BMOD;
    }
    f += p[N - std::max(s - N, 1)] * count(s, s - N, N - 1);
    if (f >= BMOD) f -= BMOD;
  }
  printf("%d\n", f % MOD);
  return 0;
}