#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10, mod = 1e9 + 7;
int rd() {
  int x = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * w;
}
int fpow(int a, int b) {
  int an = 1;
  while (b) {
    if (b & 1) an = 1ll * an * a % mod;
    a = 1ll * a * a % mod, b >>= 1;
  }
  return an;
}
int ginv(int a) { return fpow(a, mod - 2); }
int n, k, ik, pw[N << 1], prm[N], tt, mu[N], sf[N << 1], st[N << 1], tp;
bool v[N];
int main() {
  n = rd(), k = rd();
  ik = ginv(k);
  pw[0] = 1;
  for (int i = 1; i <= 2 * n; ++i) pw[i] = 1ll * pw[i - 1] * ik % mod;
  mu[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!v[i]) mu[i] = mod - 1, prm[++tt] = i;
    for (int j = 1; j <= tt && i * prm[j] <= n; ++j) {
      v[i * prm[j]] = 1;
      if (mu[i]) mu[i * prm[j]] = mod - mu[i];
      if (i % prm[j] == 0) {
        mu[i * prm[j]] = 0;
        break;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i < n; ++i) {
    int lm = (n - 1) / i;
    tp = 0;
    for (int j = 2; j <= 2 * lm; ++j)
      st[++tp] =
          1ll * pw[2 * n - i * j] * (min(lm, j - 1) - max(1, j - lm) + 1) % mod;
    for (int j = tp - 1; j; --j) st[j] = (st[j] + st[j + 1]) % mod;
    sf[0] = sf[1] = st[1];
    for (int j = 1; j <= tp; ++j) sf[j + 1] = st[j];
    tp = 1, st[0] = st[1] = 0;
    for (int j = 2; j <= 2 * lm; ++j)
      st[++tp] = min(lm, j - 1) - max(1, j - lm) + 1;
    for (int j = 1; j <= tp; ++j) st[j] = (st[j] + st[j - 1]) % mod;
    for (int j = 1; j <= i; ++j) {
      j = i / (i / j);
      if (i % j == 0 && mu[j]) {
        int l = i / j, dv = (l + n + i - 1) / i, sm;
        sm = (sf[dv] + 1ll * pw[n - l] * st[dv - 1] % mod) % mod;
        ans = (ans + 1ll * sm * mu[j]) % mod;
      }
    }
    memset(sf, 0, sizeof(int) * (tp + 5));
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate 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 = 1e5 + 10, mod = 1e9 + 7;
int rd() {
  int x = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * w;
}
int fpow(int a, int b) {
  int an = 1;
  while (b) {
    if (b & 1) an = 1ll * an * a % mod;
    a = 1ll * a * a % mod, b >>= 1;
  }
  return an;
}
int ginv(int a) { return fpow(a, mod - 2); }
int n, k, ik, pw[N << 1], prm[N], tt, mu[N], sf[N << 1], st[N << 1], tp;
bool v[N];
int main() {
  n = rd(), k = rd();
  ik = ginv(k);
  pw[0] = 1;
  for (int i = 1; i <= 2 * n; ++i) pw[i] = 1ll * pw[i - 1] * ik % mod;
  mu[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!v[i]) mu[i] = mod - 1, prm[++tt] = i;
    for (int j = 1; j <= tt && i * prm[j] <= n; ++j) {
      v[i * prm[j]] = 1;
      if (mu[i]) mu[i * prm[j]] = mod - mu[i];
      if (i % prm[j] == 0) {
        mu[i * prm[j]] = 0;
        break;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i < n; ++i) {
    int lm = (n - 1) / i;
    tp = 0;
    for (int j = 2; j <= 2 * lm; ++j)
      st[++tp] =
          1ll * pw[2 * n - i * j] * (min(lm, j - 1) - max(1, j - lm) + 1) % mod;
    for (int j = tp - 1; j; --j) st[j] = (st[j] + st[j + 1]) % mod;
    sf[0] = sf[1] = st[1];
    for (int j = 1; j <= tp; ++j) sf[j + 1] = st[j];
    tp = 1, st[0] = st[1] = 0;
    for (int j = 2; j <= 2 * lm; ++j)
      st[++tp] = min(lm, j - 1) - max(1, j - lm) + 1;
    for (int j = 1; j <= tp; ++j) st[j] = (st[j] + st[j - 1]) % mod;
    for (int j = 1; j <= i; ++j) {
      j = i / (i / j);
      if (i % j == 0 && mu[j]) {
        int l = i / j, dv = (l + n + i - 1) / i, sm;
        sm = (sf[dv] + 1ll * pw[n - l] * st[dv - 1] % mod) % mod;
        ans = (ans + 1ll * sm * mu[j]) % mod;
      }
    }
    memset(sf, 0, sizeof(int) * (tp + 5));
  }
  printf("%d\n", ans);
  return 0;
}

```