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 mod = 1000000007;
inline int add(int a, int b) {
  if ((a += b) >= mod) a -= mod;
  return a;
}
inline int mult(int a, int b) {
  long long t = 1ll * a * b;
  if (t >= mod) t %= mod;
  return t;
}
inline int dec(int a, int b) {
  if ((a -= b) < 0) a += mod;
  return a;
}
inline int power(int a, int b) {
  if (b < 0) {
    a = power(a, mod - 2);
    b = -b;
  }
  int out = 1;
  while (b) {
    if (b & 1) out = mult(out, a);
    a = mult(a, a);
    b >>= 1;
  }
  return out;
}
int n, k;
vector<int> prime;
bool flag[200010];
void init() {
  for (int i = 2; i <= 200000; i++) {
    if (!flag[i]) prime.push_back(i);
    for (int j = 0; j < prime.size() && i * prime[j] <= 200000; j++) {
      flag[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
vector<int> v[200010][2], prm[200010];
int main() {
  init();
  scanf("%d%d", &n, &k);
  int ans = 0, ans2 = 0;
  for (int i = 1; i <= 2 * n; i++) {
    int tmp = i;
    for (int j = 0; prime[j] * prime[j] <= i; j++) {
      if (tmp % prime[j] == 0) {
        prm[i].push_back(prime[j]);
        while (tmp % prime[j] == 0) tmp /= prime[j];
      }
    }
    if (tmp != 1) prm[i].push_back(tmp);
    int cnt = prm[i].size();
    for (int j = 0; j < (1 << cnt); j++) {
      tmp = 1;
      for (int k = 0; k < cnt; k++)
        if ((j >> k) & 1) tmp *= prm[i][k];
      if (__builtin_popcount(j) & 1)
        v[i][0].push_back(tmp);
      else
        v[i][1].push_back(tmp);
    }
  }
  for (int i = 1; i <= 2 * n; i++) {
    for (int j = 1; j * j <= i; j++) {
      if (i % j == 0) {
        int gcd = j, s2 = i / gcd, mx = (n - 1) / gcd;
        if (gcd >= n) continue;
        int l = max(1, s2 - mx), r = min(s2 - 1, mx), cnt = 0;
        for (int x : v[s2][1]) cnt += (r / x) - ((l - 1) / x);
        for (int x : v[s2][0]) cnt -= (r / x) - ((l - 1) / x);
        ans2 = add(ans2, mult(cnt, power(k, max(i - n, gcd) - n)));
        if (j * j != i) {
          gcd = i / j;
          if (gcd >= n) continue;
          s2 = i / gcd;
          mx = (n - 1) / gcd;
          l = max(1, s2 - mx), r = min(s2 - 1, mx);
          cnt = 0;
          for (int x : v[s2][1]) cnt += (r / x) - ((l - 1) / x);
          for (int x : v[s2][0]) cnt -= (r / x) - ((l - 1) / x);
          ans2 = add(ans2, mult(cnt, power(k, max(i - n, gcd) - n)));
        }
      }
    }
  }
  printf("%d\n", ans2);
  return 0;
}