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>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long double eps = 1e-7;
const int inf = 1000000010;
const long long INF = 10000000000000010LL;
const int mod = 1000000007;
const int MAXN = 200010, LOG = 20;
long long n, m, k, u, v, x, y, t, a, b, ans;
long long tav[MAXN], cnt[MAXN];
short mob[MAXN];
vector<int> D[MAXN];
long long powmod(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) res = res * a % mod;
  return res;
}
long long Get(long long g, long long ab) {
  long long res = 0;
  for (long long d : D[ab / g]) {
    long long lim = (n - 1) / g / d;
    long long sum = ab / g / d;
    if (!lim) continue;
    lim--;
    sum -= 2;
    long long tmp = 0;
    if (sum < 0 || 2 * lim < sum) continue;
    if (sum > lim) sum = 2 * lim - sum;
    tmp = sum + 1;
    res += mob[d] * tmp;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  tav[0] = mob[1] = 1;
  for (int i = 1; i < MAXN; i++) tav[i] = tav[i - 1] * k % mod;
  for (int i = 1; i < MAXN; i++)
    for (int j = i + i; j < MAXN; j += i) mob[j] -= mob[i];
  for (int i = 1; i < MAXN; i++)
    for (int j = i; j < MAXN; j += i) D[j].push_back(i);
  for (long long g = 1; g < n; g++)
    for (long long ab = 2 * g; ab <= 2 * n - 2; ab += g)
      cnt[max(g, ab - n)] += Get(g, ab);
  for (int i = 1; i < n; i++) ans = (ans + tav[i] * cnt[i]) % mod;
  ans = ans * powmod(k, mod - 1 - n) % mod;
  if (ans < 0) ans += mod;
  cout << ans << '\n';
  return 0;
}