#include <bits/stdc++.h>
using namespace std;
const int M = (int)1e9 + 7;
int add(int a, int b) {
  a += b;
  if (a >= M) return a - M;
  if (a < 0) return a + M;
  return a;
}
int mul(int a, int b) { return a * (long long)b % M; }
int pw(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) r = mul(r, a);
    a = mul(a, a);
    b /= 2;
  }
  return r;
}
int dv(int a, int b) { return mul(a, pw(b, M - 2)); }
const int N = (int)2e5 + 7;
int n, k, cnt[N];
vector<int> d[N];
int get(int l, int r, int val) { return r / val - (l - 1) / val; }
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 2; i <= 2 * n; i++) {
    if (d[i].empty()) {
      for (int j = i; j <= 2 * n; j += i) {
        d[j].push_back(i);
      }
    }
  }
  for (int g = 1; g <= n - 1; g++) {
    for (int s = g; s <= 2 * (n - 1); s += g) {
      int val = s / g, ret = 0, l = max(1, (s - n + g) / g),
          r = (min(n, s) - 1) / g;
      vector<int> v = d[val];
      int sz = (int)v.size();
      for (int mask = 0; mask < (1 << sz); mask++) {
        int cnt = 0, num = 1;
        for (int bit = 0; bit < sz; bit++) {
          if (mask & (1 << bit)) {
            num *= v[bit];
            cnt++;
          }
        }
        if (cnt & 1) {
          ret -= get(l, r, num);
        } else {
          ret += get(l, r, num);
        }
      }
      cnt[max(s - n, g)] = add(cnt[max(s - n, g)], ret);
    }
  }
  int invk = dv(1, k), val = 1, ret = 0;
  for (int i = 1; i <= n; i++) {
    val = mul(val, invk);
    ret = add(ret, mul(cnt[n - i], val));
  }
  cout << ret << "\n";
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 M = (int)1e9 + 7;
int add(int a, int b) {
  a += b;
  if (a >= M) return a - M;
  if (a < 0) return a + M;
  return a;
}
int mul(int a, int b) { return a * (long long)b % M; }
int pw(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) r = mul(r, a);
    a = mul(a, a);
    b /= 2;
  }
  return r;
}
int dv(int a, int b) { return mul(a, pw(b, M - 2)); }
const int N = (int)2e5 + 7;
int n, k, cnt[N];
vector<int> d[N];
int get(int l, int r, int val) { return r / val - (l - 1) / val; }
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 2; i <= 2 * n; i++) {
    if (d[i].empty()) {
      for (int j = i; j <= 2 * n; j += i) {
        d[j].push_back(i);
      }
    }
  }
  for (int g = 1; g <= n - 1; g++) {
    for (int s = g; s <= 2 * (n - 1); s += g) {
      int val = s / g, ret = 0, l = max(1, (s - n + g) / g),
          r = (min(n, s) - 1) / g;
      vector<int> v = d[val];
      int sz = (int)v.size();
      for (int mask = 0; mask < (1 << sz); mask++) {
        int cnt = 0, num = 1;
        for (int bit = 0; bit < sz; bit++) {
          if (mask & (1 << bit)) {
            num *= v[bit];
            cnt++;
          }
        }
        if (cnt & 1) {
          ret -= get(l, r, num);
        } else {
          ret += get(l, r, num);
        }
      }
      cnt[max(s - n, g)] = add(cnt[max(s - n, g)], ret);
    }
  }
  int invk = dv(1, k), val = 1, ret = 0;
  for (int i = 1; i <= n; i++) {
    val = mul(val, invk);
    ret = add(ret, mul(cnt[n - i], val));
  }
  cout << ret << "\n";
}

```