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 INF = 0x3c3c3c3c;
const long long LINF = 1ll * INF * INF * 2;
const int N = 200001;
const int MOD = 1000000007;
int rev(int x) {
  if (x == 1) return 1;
  x = x % MOD;
  if (x < 0) x += MOD;
  return ((int)(1ll * (((int)(1ll * ((MOD - 1)) * ((MOD / x)) % MOD))) *
                (rev(MOD % x)) % MOD));
}
int n, k;
int pw[N];
int phi[N];
int mp[N];
void make() {
  pw[0] = 1;
  for (int i = 1; i <= n; i++) pw[i] = ((int)(1ll * (pw[i - 1]) * (k) % MOD));
  for (int i = 2; i * i < N; i++) {
    if (mp[i]) continue;
    for (int j = i; j < N; j += i) {
      if (!mp[j]) mp[j] = i;
    }
  }
  for (int i = 2; i < N; i++)
    if (!mp[i]) mp[i] = i;
  phi[1] = 1;
  for (int i = 2; i < N; i++) {
    int t = i / mp[i];
    if (t % mp[i])
      phi[i] = ((int)(1ll * (phi[i / mp[i]]) * (mp[i] - 1) % MOD));
    else
      phi[i] = ((int)(1ll * (phi[i / mp[i]]) * (mp[i]) % MOD));
  }
}
void findD(int x, vector<int> &v) {
  v.clear();
  v.push_back(1);
  while (x > 1) {
    int p = mp[x], cnt = 0;
    while (!(x % p)) x /= p, cnt++;
    int s = ((int)((v).size()));
    for (int i = 0; i < s; i++) {
      int cur = v[i];
      for (int j = 0; j < cnt; j++) {
        cur = ((int)(1ll * (cur) * (p) % MOD));
        v.push_back(cur);
      }
    }
  }
  sort((v).begin(), (v).end());
}
void count(int l, int r, vector<int> &v, vector<int> &c) {
  c.resize(((int)((v).size())));
  for (int i = 0; i < ((int)((v).size())); i++) {
    int d = v[i];
    c[i] = r / d - (l - 1) / d;
  }
  for (int i = ((int)((v).size())) - 1; i >= 0; i--) {
    for (int j = i + 1; j < ((int)((v).size())); j++) {
      if (!(v[j] % v[i])) c[i] -= c[j];
    }
  }
}
int main(void) {
  scanf("%d%d", &n, &k);
  make();
  long long ans = 0;
  for (int i = 1; i <= n - 1; i++) ans += pw[i];
  for (int xy = 2; xy <= n; xy++) {
    vector<int> v;
    findD(xy, v);
    long long temp = 0;
    for (int d : v) temp += 1ll * phi[xy / d] / 2 * 2 * pw[d];
    ans += temp % MOD;
  }
  for (int k = 1; k < n; k++) {
    vector<int> v, c;
    findD(n + k, v);
    count(k + 1, n - 1, v, c);
    long long temp = 0;
    for (int i = 0; i < ((int)((v).size())); i++) {
      int d = v[i];
      int cnt = c[i];
      if (d > k)
        temp += 1ll * pw[d] * cnt;
      else
        temp += 1ll * pw[k] * cnt;
    }
    if (!((n + k) % 2)) temp -= 1ll * pw[(n + k) / 2];
    ans += temp % MOD;
  }
  ans %= MOD;
  ans = ((int)(1ll * (ans) * (rev(pw[n])) % MOD));
  cout << ans;
}