#include <bits/stdc++.h>
using namespace std;
const int N = 100100;
const int mod = 1000000007;
int n, K, phi[N], pr[N], cnt;
long long ans[N], pre[N];
int ksm(long long a, int b, int c = 1) {
  for (; b; b /= 2, a = a * a % mod)
    if (b & 1) c = c * a % mod;
  return c;
}
void Init() {
  phi[1] = pre[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!phi[i]) {
      pr[++cnt] = i;
      phi[i] = i - 1;
    }
    pre[i] = pre[i - 1] + phi[i];
    for (int j = 1; j <= cnt && i * pr[j] <= n; ++j)
      if (i % pr[j] == 0) {
        phi[i * pr[j]] = phi[i] * pr[j];
        break;
      } else
        phi[i * pr[j]] = phi[i] * phi[pr[j]];
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> K;
  Init();
  for (int i = 1; i < n; ++i) {
    ans[i] += pre[(n + i - 1) / i] - 1 + (n - i - 1);
    int u = (n + i - 1) / i, v = u * i - n;
    if (v) ans[v] -= phi[u];
  }
  int ret = 0;
  for (int i = n; i >= 0; --i) ret = ((long long)ret * K + ans[i]) % mod;
  ret = ksm(K, mod - 1 - n, ret);
  cout << ret << '\n';
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
const int N = 100100;
const int mod = 1000000007;
int n, K, phi[N], pr[N], cnt;
long long ans[N], pre[N];
int ksm(long long a, int b, int c = 1) {
  for (; b; b /= 2, a = a * a % mod)
    if (b & 1) c = c * a % mod;
  return c;
}
void Init() {
  phi[1] = pre[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!phi[i]) {
      pr[++cnt] = i;
      phi[i] = i - 1;
    }
    pre[i] = pre[i - 1] + phi[i];
    for (int j = 1; j <= cnt && i * pr[j] <= n; ++j)
      if (i % pr[j] == 0) {
        phi[i * pr[j]] = phi[i] * pr[j];
        break;
      } else
        phi[i * pr[j]] = phi[i] * phi[pr[j]];
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> K;
  Init();
  for (int i = 1; i < n; ++i) {
    ans[i] += pre[(n + i - 1) / i] - 1 + (n - i - 1);
    int u = (n + i - 1) / i, v = u * i - n;
    if (v) ans[v] -= phi[u];
  }
  int ret = 0;
  for (int i = n; i >= 0; --i) ret = ((long long)ret * K + ans[i]) % mod;
  ret = ksm(K, mod - 1 - n, ret);
  cout << ret << '\n';
  return 0;
}

```