#include <bits/stdc++.h>
using namespace std;
const int N = 100003, mod = 1e9 + 7;
int ksm(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = (long long)a * a % mod)
    if (b & 1) res = (long long)res * a % mod;
  return res;
}
void qmo(int &x) { x += x >> 31 & mod; }
int n, k, a[N << 1], b[N << 1], pw[N], mu[N], pri[N], tot, ans;
bool notp[N];
vector<int> vec[N];
int main() {
  scanf("%d%d", &n, &k);
  notp[0] = notp[1] = true;
  mu[1] = pw[0] = 1;
  for (int i = 1; i <= n; ++i) pw[i] = (long long)pw[i - 1] * k % mod;
  for (int i = n; i <= (n - 1 << 1); ++i) a[i] = (n << 1) - i - 1;
  for (int i = 2; i <= n; ++i) {
    if (!notp[i]) {
      pri[tot++] = i;
      mu[i] = mod - 1;
    }
    for (int j = 0; j < tot && i * pri[j] <= n; ++j) {
      notp[i * pri[j]] = true;
      if (i % pri[j])
        mu[i * pri[j]] = mod - mu[i];
      else
        break;
    }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; j += i) vec[j].emplace_back(i);
  for (int p = 1; p < n; ++p) {
    int l = (n - 1 << 1) / p;
    memset(b + 1, 0, l + 1 << 2);
    for (int d : vec[p]) {
      int m = (n + d - 1) / p;
      qmo(b[m] += mu[p / d] - mod);
      ans = (ans + (m * (m - 1ll) >> 1) % mod * pw[d] % mod * mu[p / d]) % mod;
    }
    for (int i = l - 1; i; --i) qmo(b[i] += b[i + 1] - mod);
    for (int i = 1; i <= l; ++i) qmo(a[i * p] -= b[i] * (i - 1ll) % mod);
  }
  for (int i = n; i <= (n - 1 << 1); ++i)
    ans = (ans + (long long)a[i] * pw[i - n]) % mod;
  printf("%lld\n", (long long)ans * ksm(k, mod - 1 - n) % mod);
}

### Prompt

Please formulate a cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 100003, mod = 1e9 + 7;
int ksm(int a, int b) {
  int res = 1;
  for (; b; b >>= 1, a = (long long)a * a % mod)
    if (b & 1) res = (long long)res * a % mod;
  return res;
}
void qmo(int &x) { x += x >> 31 & mod; }
int n, k, a[N << 1], b[N << 1], pw[N], mu[N], pri[N], tot, ans;
bool notp[N];
vector<int> vec[N];
int main() {
  scanf("%d%d", &n, &k);
  notp[0] = notp[1] = true;
  mu[1] = pw[0] = 1;
  for (int i = 1; i <= n; ++i) pw[i] = (long long)pw[i - 1] * k % mod;
  for (int i = n; i <= (n - 1 << 1); ++i) a[i] = (n << 1) - i - 1;
  for (int i = 2; i <= n; ++i) {
    if (!notp[i]) {
      pri[tot++] = i;
      mu[i] = mod - 1;
    }
    for (int j = 0; j < tot && i * pri[j] <= n; ++j) {
      notp[i * pri[j]] = true;
      if (i % pri[j])
        mu[i * pri[j]] = mod - mu[i];
      else
        break;
    }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; j += i) vec[j].emplace_back(i);
  for (int p = 1; p < n; ++p) {
    int l = (n - 1 << 1) / p;
    memset(b + 1, 0, l + 1 << 2);
    for (int d : vec[p]) {
      int m = (n + d - 1) / p;
      qmo(b[m] += mu[p / d] - mod);
      ans = (ans + (m * (m - 1ll) >> 1) % mod * pw[d] % mod * mu[p / d]) % mod;
    }
    for (int i = l - 1; i; --i) qmo(b[i] += b[i + 1] - mod);
    for (int i = 1; i <= l; ++i) qmo(a[i * p] -= b[i] * (i - 1ll) % mod);
  }
  for (int i = n; i <= (n - 1 << 1); ++i)
    ans = (ans + (long long)a[i] * pw[i - n]) % mod;
  printf("%lld\n", (long long)ans * ksm(k, mod - 1 - n) % mod);
}

```