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;
const int maxn = 200005;
const long long mod = 1e9 + 7;
template <typename T>
void read(T &x) {
  T flag = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') flag = -1;
  for (x = 0; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  x *= flag;
}
long long n, k;
long long mu[maxn];
vector<long long> fac[maxn];
long long prime[maxn], cnt;
bool mark[maxn];
long long ksm(long long a, long long b) {
  long long ret = 1;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) ret = ret * a % mod;
  return ret;
}
void sieve(long long m) {
  mu[1] = 1;
  for (int i = 2; i <= m; i++) {
    if (!mark[i]) prime[++cnt] = i, mu[i] = -1;
    for (int j = 1; j <= cnt && prime[j] * i <= m; j++) {
      mark[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      } else {
        mu[i * prime[j]] = -mu[i];
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    for (int j = i; j <= m; j += i) {
      fac[j].push_back(i);
    }
  }
}
long long pw[maxn];
long long cal(long long l, long long r, long long m) {
  if (l > r) return 0;
  long long ret = 0;
  for (auto i : fac[m]) {
    ret = (ret + mu[i] * (r / i - (l - 1) / i) % mod + mod) % mod;
  }
  return ret;
}
long long ans;
int main() {
  read(n);
  read(k);
  sieve(2 * n);
  pw[0] = 1;
  for (int i = 1; i <= 2 * n; i++) pw[i] = pw[i - 1] * k % mod;
  for (long long s = 2; s <= 2 * n - 2; s++) {
    for (auto i : fac[s]) {
      long long val = cal(max(1ll, s / i - (n - 1) / i),
                          min((n - 1) / i, s / i - 1), s / i);
      ans = (ans + val * pw[max(i, s - n) % mod]) % mod;
    }
  }
  printf("%lld\n", ans * ksm(pw[n], mod - 2) % mod);
  return 0;
}