#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int maxn = 2e5 + 79;
long long pwr(const long long& a, const long long& b) {
  if (!b) return 1;
  long long h = pwr(a, b >> 1);
  h = (h * h) % mod;
  if (b & 1) h = (h * a) % mod;
  return h;
}
long long inv(const long long& a, const long long& b) {
  return pwr(pwr(a, mod - 2), -b);
}
vector<int> pr, lp(maxn, 1);
int count_coprime(int n, int d) {
  if (n <= 0) return 0;
  int ans = 0, d2 = d;
  vector<int> p;
  while (d2 > 1) {
    if (p.empty() || p.back() != lp[d2]) p.push_back(lp[d2]);
    d2 /= lp[d2];
  }
  for (int i = 0; i < (1 << p.size()); i++) {
    int num = 1, cnt = 0;
    for (int j = 0; j < p.size(); j++)
      if ((1 << j) & i) num *= p[j], cnt++;
    ans += ((cnt & 1) ? -1 : 1) * (n / num);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 2; i < maxn; i++) {
    if (lp[i] == 1) lp[i] = i, pr.push_back(i);
    for (int j = 0; j < pr.size() && pr[j] <= lp[i] && pr[j] * i < maxn; j++)
      lp[pr[j] * i] = pr[j];
  }
  int n, k;
  cin >> n >> k;
  long long ans = 0;
  for (long long i = 1; i < n; i++) {
    long long e = inv(k, -i);
    ans += e;
    ans %= mod;
  }
  for (int gcd = 1; gcd <= n; gcd++) {
    for (int s = gcd; s <= 2 * n; s += gcd) {
      int s2 = s / gcd;
      int l = max(1, s2 - (n - 1) / gcd);
      int r = min(s2 - 1, (n - 1) / gcd);
      if (l > r) continue;
      long long cnt = count_coprime(r, s2) - count_coprime(l - 1, s2);
      if (s2 % 2 == 0 && s2 / 2 == 1) cnt--;
      long long e = inv(k, max(gcd, s - n) - n);
      ans += (cnt * e) % mod;
      ans %= mod;
    }
  }
  cout << ans << "\n";
  return 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 long long mod = 1e9 + 7;
const int maxn = 2e5 + 79;
long long pwr(const long long& a, const long long& b) {
  if (!b) return 1;
  long long h = pwr(a, b >> 1);
  h = (h * h) % mod;
  if (b & 1) h = (h * a) % mod;
  return h;
}
long long inv(const long long& a, const long long& b) {
  return pwr(pwr(a, mod - 2), -b);
}
vector<int> pr, lp(maxn, 1);
int count_coprime(int n, int d) {
  if (n <= 0) return 0;
  int ans = 0, d2 = d;
  vector<int> p;
  while (d2 > 1) {
    if (p.empty() || p.back() != lp[d2]) p.push_back(lp[d2]);
    d2 /= lp[d2];
  }
  for (int i = 0; i < (1 << p.size()); i++) {
    int num = 1, cnt = 0;
    for (int j = 0; j < p.size(); j++)
      if ((1 << j) & i) num *= p[j], cnt++;
    ans += ((cnt & 1) ? -1 : 1) * (n / num);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 2; i < maxn; i++) {
    if (lp[i] == 1) lp[i] = i, pr.push_back(i);
    for (int j = 0; j < pr.size() && pr[j] <= lp[i] && pr[j] * i < maxn; j++)
      lp[pr[j] * i] = pr[j];
  }
  int n, k;
  cin >> n >> k;
  long long ans = 0;
  for (long long i = 1; i < n; i++) {
    long long e = inv(k, -i);
    ans += e;
    ans %= mod;
  }
  for (int gcd = 1; gcd <= n; gcd++) {
    for (int s = gcd; s <= 2 * n; s += gcd) {
      int s2 = s / gcd;
      int l = max(1, s2 - (n - 1) / gcd);
      int r = min(s2 - 1, (n - 1) / gcd);
      if (l > r) continue;
      long long cnt = count_coprime(r, s2) - count_coprime(l - 1, s2);
      if (s2 % 2 == 0 && s2 / 2 == 1) cnt--;
      long long e = inv(k, max(gcd, s - n) - n);
      ans += (cnt * e) % mod;
      ans %= mod;
    }
  }
  cout << ans << "\n";
  return 0;
}

```