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 = 222222, mod = 1000000007;
template <typename T>
inline void read(T &x) {
  x = 0;
  bool f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  if (f) x = -x;
}
inline int qpow(int a, int b) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
int n, k, ans, pw[maxn], pr[maxn], pl, mu[maxn], s[maxn];
bool vis[maxn];
vector<int> d[maxn];
inline int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }
inline int C2(int x) { return 1ll * x * (x - 1) / 2 % mod; }
int main() {
  read(n);
  read(k);
  if (k == 1) return printf("%lld\n", 1ll * (n - 1) * (n - 1) % mod), 0;
  pw[0] = 1;
  for (int i = (1); i <= (n); i++) pw[i] = 1ll * pw[i - 1] * k % mod;
  mu[1] = 1;
  for (int i = (2); i <= (n); i++) {
    if (!vis[i]) pr[++pl] = i, mu[i] = -1;
    for (int j = (1); j <= (pl); j++) {
      int k = i * pr[j];
      if (k > n) break;
      vis[k] = true;
      if (i % pr[j] == 0) {
        mu[k] = 0;
        break;
      } else
        mu[k] = -mu[i];
    }
  }
  for (int i = (1); i <= (n); i++) mu[i] = (mod + mu[i]) % mod;
  for (int i = (1); i <= (n - 1); i++)
    for (int j = (1); j <= ((n - 1) / i); j++) {
      d[i * j].push_back(i);
      ans = (ans + 1ll * mu[j] * pw[i] % mod * C2((n + i - 1) / (i * j))) % mod;
    }
  for (int i = (1); i <= (n - 1); i++) {
    int upr = (2 * n - 2) / i;
    for (int j = (1); j <= (upr); j++) {
      if (i * j >= n)
        s[j] = 1ll * ((n - 1) / i * 2 - j + 1) * pw[i * j - n] % mod;
      else
        s[j] = 0;
    }
    s[upr + 1] = 0;
    for (int j = (upr); j >= (1); j--) s[j] = (s[j] + s[j + 1]) % mod;
    for (int j = (0); j <= ((int)d[i].size() - 1); j++)
      ans = (ans + 1ll * s[(n + d[i][j] - 1) / i + 1] * mu[i / d[i][j]]) % mod;
  }
  ans = 1ll * ans * qpow(k, mod - 1 - n) % mod;
  printf("%d\n", ans);
}