#include <bits/stdc++.h>
using namespace std;
template <int MOD_>
struct modnum {
  static constexpr int MOD = MOD_;
  static_assert(MOD_ > 0, "MOD must be positive");

 private:
  using ll = long long;
  int v;
  static int minv(int a, int m) {
    a %= m;
    assert(a);
    return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
  }

 public:
  modnum() : v(0) {}
  modnum(ll v_) : v(int(v_ % MOD)) {
    if (v < 0) v += MOD;
  }
  explicit operator int() const { return v; }
  friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
    return out << int(n);
  }
  friend std::istream& operator>>(std::istream& in, modnum& n) {
    ll v_;
    in >> v_;
    n = modnum(v_);
    return in;
  }
  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 inv() const {
    modnum res;
    res.v = minv(v, MOD);
    return res;
  }
  modnum neg() const {
    modnum res;
    res.v = v ? MOD - v : 0;
    return res;
  }
  modnum operator-() const { return neg(); }
  modnum operator+() const { return modnum(*this); }
  modnum& operator++() {
    v++;
    if (v == MOD) v = 0;
    return *this;
  }
  modnum& operator--() {
    if (v == 0) v = MOD;
    v--;
    return *this;
  }
  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.inv(); }
  friend modnum operator++(modnum& a, int) {
    modnum r = a;
    ++a;
    return r;
  }
  friend modnum operator--(modnum& a, int) {
    modnum r = a;
    --a;
    return r;
  }
  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;
  }
};
template <typename T>
T pow(T a, long long b) {
  assert(b >= 0);
  T r = 1;
  while (b) {
    if (b & 1) r *= a;
    b >>= 1;
    a *= a;
  }
  return r;
}
using num = modnum<int(1e9) + 7>;
const int MAXN = 1.1e5;
int N;
num K;
num phi[MAXN];
num phiSum[MAXN];
num kCoeff[MAXN * 2];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> N >> K;
  for (int i = 0; i <= N; i++) {
    phi[i] = i;
  }
  for (int p = 2; p <= N; p++) {
    if (phi[p] != p) continue;
    for (int j = p; j <= N; j += p) {
      phi[j] -= phi[j] / p;
    }
  }
  phiSum[0] = 0;
  for (int i = 1; i <= N; i++) {
    phiSum[i] = phiSum[i - 1] + phi[i];
  }
  for (int i = 0; i <= N - 2; i++) {
    kCoeff[i] += N - 1 - i;
  }
  for (int g = 1; g <= N - 1; g++) {
    int v = (N + g - 1) / g;
    if (v > 1) {
      kCoeff[v * g - N] -= phi[v];
      kCoeff[g] += phi[v];
    }
    kCoeff[g] += phiSum[(N - 1) / g] - 1;
  }
  num ans = 0;
  num pK = 1;
  for (int i = 0; i <= N - 1; i++, pK *= K) {
    ans += kCoeff[i] * pK;
  }
  ans /= pow(K, N);
  cout << ans << '\n';
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <int MOD_>
struct modnum {
  static constexpr int MOD = MOD_;
  static_assert(MOD_ > 0, "MOD must be positive");

 private:
  using ll = long long;
  int v;
  static int minv(int a, int m) {
    a %= m;
    assert(a);
    return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
  }

 public:
  modnum() : v(0) {}
  modnum(ll v_) : v(int(v_ % MOD)) {
    if (v < 0) v += MOD;
  }
  explicit operator int() const { return v; }
  friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
    return out << int(n);
  }
  friend std::istream& operator>>(std::istream& in, modnum& n) {
    ll v_;
    in >> v_;
    n = modnum(v_);
    return in;
  }
  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 inv() const {
    modnum res;
    res.v = minv(v, MOD);
    return res;
  }
  modnum neg() const {
    modnum res;
    res.v = v ? MOD - v : 0;
    return res;
  }
  modnum operator-() const { return neg(); }
  modnum operator+() const { return modnum(*this); }
  modnum& operator++() {
    v++;
    if (v == MOD) v = 0;
    return *this;
  }
  modnum& operator--() {
    if (v == 0) v = MOD;
    v--;
    return *this;
  }
  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.inv(); }
  friend modnum operator++(modnum& a, int) {
    modnum r = a;
    ++a;
    return r;
  }
  friend modnum operator--(modnum& a, int) {
    modnum r = a;
    --a;
    return r;
  }
  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;
  }
};
template <typename T>
T pow(T a, long long b) {
  assert(b >= 0);
  T r = 1;
  while (b) {
    if (b & 1) r *= a;
    b >>= 1;
    a *= a;
  }
  return r;
}
using num = modnum<int(1e9) + 7>;
const int MAXN = 1.1e5;
int N;
num K;
num phi[MAXN];
num phiSum[MAXN];
num kCoeff[MAXN * 2];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> N >> K;
  for (int i = 0; i <= N; i++) {
    phi[i] = i;
  }
  for (int p = 2; p <= N; p++) {
    if (phi[p] != p) continue;
    for (int j = p; j <= N; j += p) {
      phi[j] -= phi[j] / p;
    }
  }
  phiSum[0] = 0;
  for (int i = 1; i <= N; i++) {
    phiSum[i] = phiSum[i - 1] + phi[i];
  }
  for (int i = 0; i <= N - 2; i++) {
    kCoeff[i] += N - 1 - i;
  }
  for (int g = 1; g <= N - 1; g++) {
    int v = (N + g - 1) / g;
    if (v > 1) {
      kCoeff[v * g - N] -= phi[v];
      kCoeff[g] += phi[v];
    }
    kCoeff[g] += phiSum[(N - 1) / g] - 1;
  }
  num ans = 0;
  num pK = 1;
  for (int i = 0; i <= N - 1; i++, pK *= K) {
    ans += kCoeff[i] * pK;
  }
  ans /= pow(K, N);
  cout << ans << '\n';
  return 0;
}

```