#include <bits/stdc++.h>
using namespace std;
const long long N = 100005, M = 1e9 + 7;
long long n, k, ans[N], sum[N], phi[N], b[N / 10], flag[N];
inline long long ksm(long long x, long long y) {
  long long ans = 1;
  for (; y; y >>= 1, x = x * x % M)
    if (y & 1) (ans *= x) %= M;
  return ans;
}
signed main() {
  scanf("%lld%lld", &n, &k);
  phi[1] = 1;
  for (long long i = 2; i <= n; i++) {
    if (!flag[i]) {
      phi[i] = i - 1;
      b[++b[0]] = i;
    }
    for (long long j = 1; j <= b[0] && i * b[j] <= n; j++) {
      long long k = i * b[j];
      flag[k] = 1;
      if (i % b[j] == 0) {
        phi[k] = phi[i] * b[j];
        break;
      } else
        phi[k] = phi[i] * (b[j] - 1);
    }
  }
  for (long long i = 2; i <= n; i++) sum[i] = sum[i - 1] + phi[i];
  for (long long i = 0; i < n; i++) ans[i] = n - i - 1;
  for (long long d = 1; d < n; d++) {
    long long x = (n + d - 1) / d;
    ans[d] += sum[x];
    ans[x * d - n] -= phi[x];
  }
  long long tot = 0, dq = 1;
  for (long long i = 0; i < n; i++) {
    tot = (tot + ans[i] % M * dq) % M;
    dq = dq * k % M;
  }
  printf("%lld\n", tot * ksm(k, M - 1 - n) % M);
}

### 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 long long N = 100005, M = 1e9 + 7;
long long n, k, ans[N], sum[N], phi[N], b[N / 10], flag[N];
inline long long ksm(long long x, long long y) {
  long long ans = 1;
  for (; y; y >>= 1, x = x * x % M)
    if (y & 1) (ans *= x) %= M;
  return ans;
}
signed main() {
  scanf("%lld%lld", &n, &k);
  phi[1] = 1;
  for (long long i = 2; i <= n; i++) {
    if (!flag[i]) {
      phi[i] = i - 1;
      b[++b[0]] = i;
    }
    for (long long j = 1; j <= b[0] && i * b[j] <= n; j++) {
      long long k = i * b[j];
      flag[k] = 1;
      if (i % b[j] == 0) {
        phi[k] = phi[i] * b[j];
        break;
      } else
        phi[k] = phi[i] * (b[j] - 1);
    }
  }
  for (long long i = 2; i <= n; i++) sum[i] = sum[i - 1] + phi[i];
  for (long long i = 0; i < n; i++) ans[i] = n - i - 1;
  for (long long d = 1; d < n; d++) {
    long long x = (n + d - 1) / d;
    ans[d] += sum[x];
    ans[x * d - n] -= phi[x];
  }
  long long tot = 0, dq = 1;
  for (long long i = 0; i < n; i++) {
    tot = (tot + ans[i] % M * dq) % M;
    dq = dq * k % M;
  }
  printf("%lld\n", tot * ksm(k, M - 1 - n) % M);
}

```