#include <bits/stdc++.h>
const int mod = 1e9 + 7;
int qpow(int a, int b) {
  if (b == 0) return 1;
  long long d = qpow(a, b >> 1);
  d = d * d % mod;
  if (b & 1) d = d * a % mod;
  return d;
}
int inv(int x) {
  return x == 1 ? 1 : (long long)(mod - mod / x) * inv(mod % x) % mod;
}
int mu[100005], prime[100005], tl, not_prime[100005], n, m, ans, pw[100005],
    pt[100005], I;
void sieve(int lim) {
  pw[0] = 1;
  for (int i = 1; i <= lim; ++i) pw[i] = (long long)pw[i - 1] * m % mod;
  mu[1] = 1;
  for (int i = 2; i <= lim; ++i) {
    if (!not_prime[i]) {
      prime[++tl] = i;
      mu[i] = mod - 1;
    }
    for (int j = 1; j <= tl && i * prime[j] <= lim; ++j) {
      not_prime[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      } else
        mu[i * prime[j]] = mod - mu[i];
    }
  }
}
int cal(int w, int n) { return (long long)(mod + 1 - pt[n + 1]) * I % mod; }
int calc(int w, int l, int r) {
  if (l > r) return 0;
  if (w == 1) return (r - l + 1);
  return (cal(w, r) - cal(w, l - 1) + mod) % mod;
}
int main() {
  scanf("%d%d", &n, &m);
  int q = inv(qpow(m, n));
  sieve(n);
  int S = n - 1;
  pt[0] = 1;
  for (int d = 1; d <= S; ++d) {
    int L = S / d;
    for (int p = 1; p <= L; ++p) {
      int M = L / p;
      int W = qpow(m, d * p);
      int p1 = M;
      int tot = 0;
      int nw = 1;
      if (W != 1) I = inv((mod + 1 - W) % mod);
      for (int x = 1; x <= M + 1; ++x) pt[x] = (long long)pt[x - 1] * W % mod;
      for (int x = 1; x <= M; ++x) {
        nw = (long long)nw * W % mod;
        while ((long long)d * p * (x + p1) - n > d && p1 > 0) p1--;
        tot = (tot + (long long)pw[d] % mod * p1 % mod) % mod;
        tot = (tot + (long long)nw * q % mod * calc(W, p1 + 1, M) % mod) % mod;
      }
      ans = (ans + (long long)tot * mu[p] % mod) % mod;
    }
  }
  printf("%lld", (long long)q * ans % mod);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
const int mod = 1e9 + 7;
int qpow(int a, int b) {
  if (b == 0) return 1;
  long long d = qpow(a, b >> 1);
  d = d * d % mod;
  if (b & 1) d = d * a % mod;
  return d;
}
int inv(int x) {
  return x == 1 ? 1 : (long long)(mod - mod / x) * inv(mod % x) % mod;
}
int mu[100005], prime[100005], tl, not_prime[100005], n, m, ans, pw[100005],
    pt[100005], I;
void sieve(int lim) {
  pw[0] = 1;
  for (int i = 1; i <= lim; ++i) pw[i] = (long long)pw[i - 1] * m % mod;
  mu[1] = 1;
  for (int i = 2; i <= lim; ++i) {
    if (!not_prime[i]) {
      prime[++tl] = i;
      mu[i] = mod - 1;
    }
    for (int j = 1; j <= tl && i * prime[j] <= lim; ++j) {
      not_prime[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      } else
        mu[i * prime[j]] = mod - mu[i];
    }
  }
}
int cal(int w, int n) { return (long long)(mod + 1 - pt[n + 1]) * I % mod; }
int calc(int w, int l, int r) {
  if (l > r) return 0;
  if (w == 1) return (r - l + 1);
  return (cal(w, r) - cal(w, l - 1) + mod) % mod;
}
int main() {
  scanf("%d%d", &n, &m);
  int q = inv(qpow(m, n));
  sieve(n);
  int S = n - 1;
  pt[0] = 1;
  for (int d = 1; d <= S; ++d) {
    int L = S / d;
    for (int p = 1; p <= L; ++p) {
      int M = L / p;
      int W = qpow(m, d * p);
      int p1 = M;
      int tot = 0;
      int nw = 1;
      if (W != 1) I = inv((mod + 1 - W) % mod);
      for (int x = 1; x <= M + 1; ++x) pt[x] = (long long)pt[x - 1] * W % mod;
      for (int x = 1; x <= M; ++x) {
        nw = (long long)nw * W % mod;
        while ((long long)d * p * (x + p1) - n > d && p1 > 0) p1--;
        tot = (tot + (long long)pw[d] % mod * p1 % mod) % mod;
        tot = (tot + (long long)nw * q % mod * calc(W, p1 + 1, M) % mod) % mod;
      }
      ans = (ans + (long long)tot * mu[p] % mod) % mod;
    }
  }
  printf("%lld", (long long)q * ans % mod);
  return 0;
}

```