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;
inline int read() {
  char ch = getchar();
  int w = 1, c = 0;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') w = -1;
  for (; isdigit(ch); ch = getchar()) c = (c << 1) + (c << 3) + (ch ^ 48);
  return w * c;
}
const int M = 2e5 + 10, mod = 1e9 + 7;
int phi[M];
bool p[M];
int P[M], cnt;
void sieve(int n) {
  for (int i = (2); i <= (n); ++i) {
    if (!p[i]) {
      P[++cnt] = i;
      phi[i] = i - 1;
    }
    for (int j = 1; j <= cnt && P[j] * i <= n; j++) {
      p[i * P[j]] = 1;
      if (i % P[j] == 0) {
        phi[i * P[j]] = phi[i] * P[j];
        break;
      }
      phi[i * P[j]] = phi[i] * phi[P[j]];
    }
  }
}
int n, m;
long long pw[M], sp[M];
long long fast(long long x, long long p) {
  long long ret = 1;
  for (; p; p >>= 1, x = x * x % mod)
    if (p & 1) ret = ret * x % mod;
  return ret;
}
int main() {
  sieve(111111);
  n = read();
  m = read();
  pw[0] = 1;
  n--;
  long long sm = 0;
  for (int i = (1); i <= (111111); ++i)
    pw[i] = pw[i - 1] * m % mod, sp[i] = (sp[i - 1] + phi[i]) % mod;
  for (int i = (1); i <= (n); ++i) sm = (sm + pw[i]) % mod;
  long long ipn = fast(pw[n + 1], mod - 2);
  long long ans = sm * sm % mod * ipn % mod;
  for (int i = (1); i <= (n); ++i) {
    ans = (ans + pw[i] * sp[n / i + 1]) % mod;
  }
  long long del = 0;
  for (int i = (1); i <= (n); ++i) {
    long long tmp = pw[i];
    for (int j = (2); j <= (n / i + 1); ++j) {
      tmp = tmp * pw[i] % mod;
      del = (del + 1ll * (mod - tmp) * phi[j]) % mod;
    }
  }
  del = del * ipn % mod;
  ans = (ans + del) % mod;
  ans = ans * ipn % mod;
  cout << ans << "\n";
  return 0;
}