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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
namespace _buff {
const size_t BUFF = 1 << 19;
char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
char getc() {
  if (ib == ie) {
    ib = ibuf;
    ie = ibuf + fread(ibuf, 1, BUFF, stdin);
  }
  return ib == ie ? -1 : *ib++;
}
}  // namespace _buff
LL read() {
  using namespace _buff;
  LL ret = 0;
  bool pos = true;
  char c = getc();
  for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
    assert(~c);
  }
  if (c == '-') {
    pos = false;
    c = getc();
  }
  for (; c >= '0' && c <= '9'; c = getc()) {
    ret = (ret << 3) + (ret << 1) + (c ^ 48);
  }
  return pos ? ret : -ret;
}
const size_t N = 2e5 + 5;
const int MOD = 1e9 + 7;
int qpow(int base, int e) {
  int ret = 1;
  for (; e; e >>= 1) {
    if (e & 1) {
      ret = (LL)ret * base % MOD;
    }
    base = (LL)base * base % MOD;
  }
  return ret;
}
int n, k, pw[N];
int mu[N];
bool npr[N];
vector<int> divs[N];
void prep() {
  vector<int> pri;
  mu[1] = 1;
  int m = n << 1;
  for (int i = 2; i <= m; ++i) {
    if (!npr[i]) {
      mu[i] = -1;
      pri.emplace_back(i);
    }
    for (int j : pri) {
      int k = i * j;
      if (k >= N) break;
      npr[k] = true;
      if (i % j == 0) {
        mu[k] = 0;
        break;
      }
      mu[k] = -mu[i];
    }
  }
  for (int i = 1; i <= m; ++i) {
    if (mu[i]) {
      for (int j = i; j <= m; j += i) {
        divs[j].emplace_back(i);
      }
    }
  }
  pw[0] = 1;
  pw[1] = qpow(k, MOD - 2);
  for (int i = 2; i <= m; ++i) {
    pw[i] = (LL)pw[i - 1] * pw[1] % MOD;
  }
}
int calc(int n, int l, int r) {
  if (l > r) return 0;
  int ret = 0;
  for (int d : divs[n]) {
    ret += mu[d] * (r / d - (l - 1) / d);
  }
  return ret;
}
int main() {
  n = read();
  k = read();
  prep();
  int ans = 0;
  for (int d = 1; d < n; ++d) {
    for (int s = 2; d * s <= 2 * (n - 1); ++s) {
      ans = (ans + (LL)pw[n - max(d * s - n, d)] *
                       calc(s, max(1, (d * s - n + d) / d),
                            min((n - 1) / d, s - 1))) %
            MOD;
    }
  }
  cout << ans << '\n';
  return 0;
}