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 N = 2e5 + 5;
const int mod = 1e9 + 7;
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int mul(int a, int b) { return 1ll * a * b % mod; }
int Pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
int inv(int a, int p) { return a == 1 ? 1 : p - 1ll * p * inv(p % a, a) / a; }
vector<int> dvs[N];
int f[N];
int c[N];
int P[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int k;
  cin >> k;
  P[0] = 1;
  f[1] = 1;
  for (int i = 1; i < N; ++i) P[i] = mul(P[i - 1], k);
  for (int i = 1; i < N; ++i)
    for (int j = i; j < N; j += i) dvs[j].emplace_back(i);
  for (int i = 2; i < N; ++i) {
    int p = dvs[i][1];
    int x = i / p;
    if (x % p == 0)
      f[i] = 0;
    else
      f[i] = -f[x];
  }
  auto upd = [&](int x, int v) {
    for (int d : dvs[x]) add(c[d], v);
  };
  auto get = [&](int x) {
    int ans = 0;
    for (int d : dvs[x]) {
      if (f[d] > 0) add(ans, c[d]);
      if (f[d] < 0) sub(ans, c[d]);
    }
    return ans;
  };
  int ans = 0;
  for (int i = 1; i < n; ++i) {
    int r = n / i + (n % i > 0);
    int S = 0;
    for (int j = 1; j < r; ++j) upd(j, 1);
    for (int j = 1; j < r; ++j) {
      add(S, get(j));
      upd(r - j, -1);
    }
    add(ans, mul(S, P[i]));
    S = 0;
    for (int j = 1; j < r; ++j) {
      add(S, mul(get(j), P[i * j]));
      upd(r - j, P[i * r - i * j]);
    }
    for (int j = 1; j < r; ++j) upd(j, mod - P[i * j]);
    add(ans, mul(S, inv(P[n], mod)));
  }
  ans = mul(ans, inv(P[n], mod));
  cout << ans << endl;
}