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 = 1e9 + 7;
const int N = 2e5 + 2;
void add(int &x, int y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
}
void sub(int &x, int y) {
  x -= y;
  if (x < 0) {
    x += mod;
  }
}
int mul(int x, int y) { return (1ll * x * y) % mod; }
int binpow(int x, int y) {
  int tich = 1;
  while (y) {
    if (y & 1) {
      tich = mul(tich, x);
    }
    x = mul(x, x);
    y >>= 1;
  }
  return tich;
}
int backtrack(vector<int> &lis, int lef, int rig, int ptr, int cnt, int val,
              int s) {
  if (ptr == lis.size()) {
    return cnt * (rig / val - (lef - 1) / val);
  }
  return backtrack(lis, lef, rig, ptr + 1, cnt, val, s) +
         backtrack(lis, lef, rig, ptr + 1, -cnt, val * lis[ptr], s);
}
int pw[N];
vector<int> uoc[N], uocnt[N];
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, i, j, k, l, num, ans = 0, r;
  cin >> n >> num;
  pw[0] = 1;
  for (i = 1; i <= n; i++) {
    pw[i] = mul(pw[i - 1], num);
  }
  for (i = 1; i <= 2 * n; i++) {
    for (j = i; j <= 2 * n; j += i) {
      uoc[j].push_back(i);
    }
    if (i == 1) {
      continue;
    }
    if (uoc[i].size() == 2) {
      for (j = i; j <= 2 * n; j += i) {
        uocnt[j].push_back(i);
      }
    }
  }
  for (int s = 3; s < 2 * n - 2; s++) {
    for (int gcd : uoc[s]) {
      int news = s / gcd, l = max(1, news - (n - 1) / gcd),
          r = min((n - 1) / gcd, news - 1);
      if (l > r) {
        continue;
      }
      j = backtrack(uocnt[news], l, r, 0, 1, 1, news);
      if (news == 2) {
        j--;
      }
      add(ans, mul(j, pw[max(s - n, gcd)]));
    }
  }
  for (i = 1; i < n; i++) {
    add(ans, pw[n - i]);
  }
  cout << mul(ans, binpow(pw[n], mod - 2));
}