#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10, mod = 1e9 + 7;
const long long inf = 1e18;
int pw[maxn], par[maxn];
int arr[100], zrb[100], C;
int Pow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod) {
    if (b & 1) ans = 1ll * ans * a % mod;
  }
  return ans;
}
int f(int n, int lm) {
  if (lm <= 0) return 0;
  C = 0;
  while (n > 1) {
    int p = par[n];
    while (n % p == 0) n /= p;
    arr[C++] = p;
  }
  int ans = lm;
  zrb[0] = 1;
  for (int i = 1; i < (1 << C); i++) {
    zrb[i] = zrb[i ^ (i & -i)] * arr[__builtin_ctz(i)];
    ans += (__builtin_popcount(i) & 1 ? -1 : 1) * (lm / zrb[i]);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie();
  int n, x;
  cin >> n >> x;
  pw[0] = 1;
  for (int i = 1; i < maxn; i++) pw[i] = 1ll * pw[i - 1] * x % mod;
  for (int i = 2; i < maxn; i++) {
    if (par[i] == 0) {
      for (int j = i; j < maxn; j += i) par[j] = i;
    }
  }
  int ans = 0;
  for (int g = 1; g < n; g++) {
    for (int sm = 2 * g; sm <= 2 * (n - 1); sm += g) {
      int lm = min((n + g - 1) / g, sm / g) - 1;
      int num = f(sm / g, lm) - f(sm / g, (sm / g) - ((n + g - 1) / g));
      ans = (1ll * ans + 1ll * num *
                             pw[(n % g) * max(1, (sm / g) - ((n + g - 1) / g)) +
                                (g - (n % g)) * max(1, (sm / g) - (n / g))]) %
            mod;
    }
  }
  ans = 1ll * ans * Pow(Pow(x, mod - 2), n) % mod;
  if (ans < 0) ans += mod;
  return cout << ans << endl, 0;
}

### Prompt

Your task is to create 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 maxn = 2e5 + 10, mod = 1e9 + 7;
const long long inf = 1e18;
int pw[maxn], par[maxn];
int arr[100], zrb[100], C;
int Pow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod) {
    if (b & 1) ans = 1ll * ans * a % mod;
  }
  return ans;
}
int f(int n, int lm) {
  if (lm <= 0) return 0;
  C = 0;
  while (n > 1) {
    int p = par[n];
    while (n % p == 0) n /= p;
    arr[C++] = p;
  }
  int ans = lm;
  zrb[0] = 1;
  for (int i = 1; i < (1 << C); i++) {
    zrb[i] = zrb[i ^ (i & -i)] * arr[__builtin_ctz(i)];
    ans += (__builtin_popcount(i) & 1 ? -1 : 1) * (lm / zrb[i]);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie();
  int n, x;
  cin >> n >> x;
  pw[0] = 1;
  for (int i = 1; i < maxn; i++) pw[i] = 1ll * pw[i - 1] * x % mod;
  for (int i = 2; i < maxn; i++) {
    if (par[i] == 0) {
      for (int j = i; j < maxn; j += i) par[j] = i;
    }
  }
  int ans = 0;
  for (int g = 1; g < n; g++) {
    for (int sm = 2 * g; sm <= 2 * (n - 1); sm += g) {
      int lm = min((n + g - 1) / g, sm / g) - 1;
      int num = f(sm / g, lm) - f(sm / g, (sm / g) - ((n + g - 1) / g));
      ans = (1ll * ans + 1ll * num *
                             pw[(n % g) * max(1, (sm / g) - ((n + g - 1) / g)) +
                                (g - (n % g)) * max(1, (sm / g) - (n / g))]) %
            mod;
    }
  }
  ans = 1ll * ans * Pow(Pow(x, mod - 2), n) % mod;
  if (ans < 0) ans += mod;
  return cout << ans << endl, 0;
}

```