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 MOD = 1000000007;
int N, K;
vector<int> Primes[200005];
vector<int> Div[200005];
void fillPrimes(int x) {
  int i = 2, init = x;
  while (i <= x && i * i <= init) {
    if (x % i == 0) Primes[init].push_back(i);
    while (x % i == 0) x /= i;
    ++i;
  }
  if (x > 1) Primes[init].push_back(x);
}
void precalcDivs() {
  for (int i = 1; i <= 2 * N - 3; i++) {
    for (int j = i; j <= 2 * N - 3; j += i) Div[j].push_back(i);
  }
}
int findComponents(int sum, int gcd) {
  int s2 = sum / gcd;
  int left = max(1, s2 - (N - 1) / gcd);
  int right = min(s2 - 1, (N - 1) / gcd);
  if (s2 <= 2) return 0;
  int ans = 0;
  for (int conf = 1; conf < (1 << Primes[s2].size()); conf++) {
    int cnt = -1, prod = 1;
    for (int i = 0; i < Primes[s2].size(); i++) {
      if (conf & (1 << i)) cnt = -cnt, prod *= Primes[s2][i];
    }
    ans += cnt * (right / prod - (left - 1) / prod);
  }
  return right - left + 1 - ans;
}
inline void Add(int& x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
int powerLog(int n, int p) {
  int sol = 1;
  while (p) {
    if (p % 2 == 1) sol = (1LL * sol * n) % MOD;
    p /= 2;
    n = (1LL * n * n) % MOD;
  }
  return sol;
}
void Solve() {
  int ans = 0;
  for (int i = 3; i <= 2 * N - 3; i++) {
    for (int j = 0; j < Div[i].size(); j++) {
      Add(ans, (1LL * powerLog(K, max(i - N, Div[i][j])) *
                findComponents(i, Div[i][j])) %
                   MOD);
    }
  }
  for (int i = 1; i < N; i++) Add(ans, powerLog(K, N - i));
  ans = (1LL * ans * powerLog(powerLog(K, N), MOD - 2)) % MOD;
  cout << ans << '\n';
}
int main() {
  cin >> N >> K;
  for (int i = 2; i <= 2 * N - 3; i++) fillPrimes(i);
  precalcDivs();
  Solve();
  return 0;
}