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;
using ll = long long;
ll modinv(ll a, ll m) {
  assert(m > 0);
  if (m == 1) return 0;
  a %= m;
  if (a < 0) a += m;
  assert(a != 0);
  if (a == 1) return 1;
  return m - modinv(m, a) * m / a;
}
template <int MOD_>
struct modnum {
 private:
  int v;

 public:
  static const int MOD = MOD_;
  modnum() : v(0) {}
  modnum(ll v_) : v(int(v_ % MOD)) {
    if (v < 0) v += MOD;
  }
  explicit operator int() const { return v; }
  friend bool operator==(const modnum& a, const modnum& b) {
    return a.v == b.v;
  }
  friend bool operator!=(const modnum& a, const modnum& b) {
    return a.v != b.v;
  }
  modnum operator~() const {
    modnum res;
    res.v = modinv(v, MOD);
    return res;
  }
  modnum& operator+=(const modnum& o) {
    v += o.v;
    if (v >= MOD) v -= MOD;
    return *this;
  }
  modnum& operator-=(const modnum& o) {
    v -= o.v;
    if (v < 0) v += MOD;
    return *this;
  }
  modnum& operator*=(const modnum& o) {
    v = int(ll(v) * ll(o.v) % MOD);
    return *this;
  }
  modnum& operator/=(const modnum& o) { return *this *= (~o); }
  friend modnum operator+(const modnum& a, const modnum& b) {
    return modnum(a) += b;
  }
  friend modnum operator-(const modnum& a, const modnum& b) {
    return modnum(a) -= b;
  }
  friend modnum operator*(const modnum& a, const modnum& b) {
    return modnum(a) *= b;
  }
  friend modnum operator/(const modnum& a, const modnum& b) {
    return modnum(a) /= b;
  }
};
using num = modnum<int(1e9) + 7>;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  vector<num> powk = {1};
  for (int i = 1; i <= 4 * n; i++) {
    powk.push_back(powk.back() * k);
  }
  vector<num> cnt(n + 1, 0);
  vector<num> phi(2 * n + 1);
  vector<num> mu(2 * n + 1);
  for (int i = 1; i <= 2 * n; i++) {
    phi[i] = i;
    mu[i] = (i == 1);
  }
  for (int i = 1; i <= 2 * n; i++) {
    for (int d = 2 * i; d <= 2 * n; d += i) {
      phi[d] -= phi[i];
      mu[d] -= mu[i];
    }
  }
  for (int g = 1; g < n; g++) {
    int rmax = (n - 1) / g;
    int sum_thresh = rmax + 1;
    vector<num> phi2(2 * rmax + 1, 0);
    for (int i = 1; i <= 2 * rmax; i++) {
      for (int d = 1; d <= 2 * rmax / i; d++) {
        phi2[d * i] += (rmax / d) * mu[d];
      }
    }
    for (int s = 2; s <= rmax * 2; s += 1) {
      if (s <= sum_thresh) {
        cnt[g] += phi[s];
      } else {
        num z = 2 * phi2[s] - phi[s];
        cnt[g * s - n] += z;
      }
    }
  }
  num tot = 0;
  for (int i = 1; i < n; i++) {
    tot += powk[i] * cnt[i];
  }
  tot /= powk[n];
  cout << (int)tot << '\n';
}