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 MOD = 1e9 + 7;
int N, K;
int prime[MAXN];
vector<pair<int, int> > rad[MAXN];
long long cnt[MAXN];
void load() { scanf("%d%d", &N, &K); }
inline int add(int x, int y) {
  x += y;
  if (x >= MOD) return x - MOD;
  if (x < 0) return x + MOD;
  return x;
}
inline int mul(int x, int y) { return (long long)x * y % MOD; }
int pot(int x, int y) {
  int res = 1;
  for (; y; y /= 2) {
    if (y % 2) res = mul(res, x);
    x = mul(x, x);
  }
  return res;
}
int calc(int x, int y) {
  if (y <= 0) return 0;
  int res = 0;
  for (auto it : rad[x]) res += y / it.first * it.second;
  return res;
}
void add(int g) {
  int n = (N - 1) / g;
  int a = N % g;
  int b = g - a;
  for (int i = 3; i <= 2 * n; i++)
    cnt[a * max(i - N / g - 1, 1) + b * max(i - N / g, 1)] +=
        calc(i, min(i, n)) - calc(i, i - n - 1);
}
int solve() {
  for (int i = 2; i <= 2 * N; i++)
    if (!prime[i])
      for (int j = i; j <= 2 * N; j += i)
        if (!prime[j]) prime[j] = i;
  for (int i = 1; i <= 2 * N; i++) {
    int x = i;
    vector<int> p;
    while (x > 1) {
      int tmp = prime[x];
      p.push_back(tmp);
      while (!(x % tmp)) x /= tmp;
    }
    int sz = p.size();
    for (int j = 0; j < 1 << sz; j++) {
      int prod = 1;
      for (int k = 0; k < sz; k++)
        if (j >> k & 1) prod *= p[k];
      rad[i].push_back({prod, __builtin_popcount(j) % 2 ? -1 : 1});
    }
  }
  for (int i = 1; i < N; i++) {
    cnt[i]++;
    add(i);
  }
  int sol = 0;
  for (int i = 1; i <= N; i++) sol = add(sol, mul(cnt[i] % MOD, pot(K, i)));
  return mul(sol, pot(K, MOD - 1 - N));
}
int main() {
  load();
  printf("%d\n", solve());
  return 0;
}