#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
inline int add(int a, int b) {
  a += b;
  return a > mod ? a - mod : a;
}
inline int sub(int a, int b) {
  a -= b;
  return a < 0 ? a + mod : a;
}
inline int mul(int a, int b) { return (long long)a * b % mod; }
inline int qpow(int a, int b) {
  int ret = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) ret = mul(ret, a);
  return ret;
}
int n, k;
const int N = 2e5 + 5;
vector<int> d[N];
long long c[N];
long long dp[N], ans[N];
int main() {
  for (int i = 1; i < N; i++)
    for (int j = i; j < N; j += i) d[j].push_back(i);
  cin >> n >> k;
  for (int s = n + 1; s < n * 2 - 1; ++s) {
    int l = s - n, r = n - 1;
    for (size_t i = 0; i < d[s].size(); i++) {
      int x = d[s][i];
      c[x] = (r / x) - (l / x);
    }
    for (size_t i = (int)d[s].size() - 1; ~i; i--) {
      int x = d[s][i];
      for (size_t j = 0; j < d[x].size(); j++) {
        int y = d[x][j];
        if (x != y) c[y] -= c[x];
      }
    }
    for (size_t i = 0; i < d[s].size(); i++) {
      int g = d[s][i];
      if (s - g >= n) ans[s - n] += c[g];
    }
  }
  for (int m = 2; m <= n; m++) {
    dp[m] = (long long)m * (m - 1) / 2;
    long long b = (long long)sqrt(m);
    while (b * b < m) b++;
    for (int i = 2; i <= b; i++) {
      dp[m] -= dp[m / i];
    }
    for (int i = b; i; i--) {
      long long z = m / i;
      if (z > b) {
        dp[m] -= (z - b) * dp[i];
        b = z;
      }
    }
  }
  for (int g = 1; g < n; g++) {
    ans[g] += dp[(n - 1) / g + 1];
  }
  int res = 0;
  for (int i = 1; i < n; i++) {
    ans[i] += mod;
    ans[i] %= mod;
    res = add(res, mul(ans[i], qpow(k, i)));
  }
  res = mul(res, qpow(qpow(k, n), mod - 2));
  cout << res << endl;
  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;
const int mod = 1e9 + 7;
inline int add(int a, int b) {
  a += b;
  return a > mod ? a - mod : a;
}
inline int sub(int a, int b) {
  a -= b;
  return a < 0 ? a + mod : a;
}
inline int mul(int a, int b) { return (long long)a * b % mod; }
inline int qpow(int a, int b) {
  int ret = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) ret = mul(ret, a);
  return ret;
}
int n, k;
const int N = 2e5 + 5;
vector<int> d[N];
long long c[N];
long long dp[N], ans[N];
int main() {
  for (int i = 1; i < N; i++)
    for (int j = i; j < N; j += i) d[j].push_back(i);
  cin >> n >> k;
  for (int s = n + 1; s < n * 2 - 1; ++s) {
    int l = s - n, r = n - 1;
    for (size_t i = 0; i < d[s].size(); i++) {
      int x = d[s][i];
      c[x] = (r / x) - (l / x);
    }
    for (size_t i = (int)d[s].size() - 1; ~i; i--) {
      int x = d[s][i];
      for (size_t j = 0; j < d[x].size(); j++) {
        int y = d[x][j];
        if (x != y) c[y] -= c[x];
      }
    }
    for (size_t i = 0; i < d[s].size(); i++) {
      int g = d[s][i];
      if (s - g >= n) ans[s - n] += c[g];
    }
  }
  for (int m = 2; m <= n; m++) {
    dp[m] = (long long)m * (m - 1) / 2;
    long long b = (long long)sqrt(m);
    while (b * b < m) b++;
    for (int i = 2; i <= b; i++) {
      dp[m] -= dp[m / i];
    }
    for (int i = b; i; i--) {
      long long z = m / i;
      if (z > b) {
        dp[m] -= (z - b) * dp[i];
        b = z;
      }
    }
  }
  for (int g = 1; g < n; g++) {
    ans[g] += dp[(n - 1) / g + 1];
  }
  int res = 0;
  for (int i = 1; i < n; i++) {
    ans[i] += mod;
    ans[i] %= mod;
    res = add(res, mul(ans[i], qpow(k, i)));
  }
  res = mul(res, qpow(qpow(k, n), mod - 2));
  cout << res << endl;
  return 0;
}

```