#include <bits/stdc++.h>
int n, k;
const int P = 1000000007;
inline int mul(const int &a, const int &b) { return 1ll * a * b % P; }
inline int add(int a, const int &b) {
  a += b;
  return (a >= P) ? a - P : a;
}
inline int sub(int a, const int &b) {
  a -= b;
  return (a < 0) ? a + P : a;
}
int f[1000001], prime[1000001], top, cnt[1000001], mu[1000001], phi[1000001];
bool pri[1000001];
int qsm(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
std::vector<int> Div[200001];
int get(int a, int b) {
  int l = std::max(0, b - a - 1), r = std::min(a, b - 1);
  if (l > r) return 0;
  if (b - 1 <= r) return b == 1 ? 0 : phi[b];
  int ans = 0;
  for (int i = 0, tem; i < Div[b].size(); ++i) {
    tem = Div[b][i];
    if (tem > r) break;
    ans = add(ans, mul(mu[tem], (r / tem) - (l / tem)));
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  k %= P;
  mu[1] = phi[1] = 1;
  for (int i = 2; i < (n << 1); i++) {
    if (!pri[i]) prime[++top] = i, mu[i] = P - 1, phi[i] = i - 1;
    for (int j = 1; j <= top && i * prime[j] < (n << 1); j++) {
      pri[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        phi[i * prime[j]] = phi[i] * prime[j];
        break;
      }
      phi[i * prime[j]] = phi[i] * (prime[j] - 1);
      mu[i * prime[j]] = P - mu[i];
    }
  }
  for (int i = 1; i < (n << 1) - 1; i++)
    for (int j = i; j < (n << 1) - 1; j += i) Div[j].push_back(i);
  k = qsm(k, P - 2);
  int ans = 0, v = 1;
  for (int i = 1; i < n; i++)
    for (int j = i; j < (n << 1) - 1; j += i)
      cnt[n - std::max(j - n, i)] =
          add(cnt[n - std::max(j - n, i)], get((n - 1) / i, j / i));
  for (int i = 0; i < n; i++) ans = add(ans, mul(cnt[i], v)), v = mul(v, k);
  printf("%d", ans);
}

### 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>
int n, k;
const int P = 1000000007;
inline int mul(const int &a, const int &b) { return 1ll * a * b % P; }
inline int add(int a, const int &b) {
  a += b;
  return (a >= P) ? a - P : a;
}
inline int sub(int a, const int &b) {
  a -= b;
  return (a < 0) ? a + P : a;
}
int f[1000001], prime[1000001], top, cnt[1000001], mu[1000001], phi[1000001];
bool pri[1000001];
int qsm(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
std::vector<int> Div[200001];
int get(int a, int b) {
  int l = std::max(0, b - a - 1), r = std::min(a, b - 1);
  if (l > r) return 0;
  if (b - 1 <= r) return b == 1 ? 0 : phi[b];
  int ans = 0;
  for (int i = 0, tem; i < Div[b].size(); ++i) {
    tem = Div[b][i];
    if (tem > r) break;
    ans = add(ans, mul(mu[tem], (r / tem) - (l / tem)));
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  k %= P;
  mu[1] = phi[1] = 1;
  for (int i = 2; i < (n << 1); i++) {
    if (!pri[i]) prime[++top] = i, mu[i] = P - 1, phi[i] = i - 1;
    for (int j = 1; j <= top && i * prime[j] < (n << 1); j++) {
      pri[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        phi[i * prime[j]] = phi[i] * prime[j];
        break;
      }
      phi[i * prime[j]] = phi[i] * (prime[j] - 1);
      mu[i * prime[j]] = P - mu[i];
    }
  }
  for (int i = 1; i < (n << 1) - 1; i++)
    for (int j = i; j < (n << 1) - 1; j += i) Div[j].push_back(i);
  k = qsm(k, P - 2);
  int ans = 0, v = 1;
  for (int i = 1; i < n; i++)
    for (int j = i; j < (n << 1) - 1; j += i)
      cnt[n - std::max(j - n, i)] =
          add(cnt[n - std::max(j - n, i)], get((n - 1) / i, j / i));
  for (int i = 0; i < n; i++) ans = add(ans, mul(cnt[i], v)), v = mul(v, k);
  printf("%d", ans);
}

```