#include <bits/stdc++.h>
using namespace std;
namespace zzc {
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
const long long maxn = 1e5 + 5;
const long long mod = 1e9 + 7;
long long n, k, ans, cnt;
long long p[maxn], phi[maxn], pw[maxn];
bool vis[maxn];
long long qpow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
void init() {
  phi[1] = 1;
  for (register int i = 2; i <= n; i++) {
    if (!vis[i]) {
      p[++cnt] = i;
      phi[i] = i - 1;
    }
    for (register int j = 1; j <= cnt && i * p[j] <= n; j++) {
      vis[i * p[j]] = true;
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      }
      phi[i * p[j]] = phi[i] * (p[j] - 1);
    }
  }
  for (register int i = 1; i <= n; i++) phi[i] = (phi[i] + phi[i - 1]) % mod;
}
void work() {
  n = read();
  k = read();
  init();
  pw[0] = 1;
  for (register int i = 1; i <= n; i++) pw[i] = pw[i - 1] * k % mod;
  for (register long long i = 1; i <= n - 2; i++)
    ans = (ans + pw[i] * (n - 1 - i) % mod);
  pw[0] = 0;
  for (register long long p = 1; p < n; p++) {
    int lim = (n - 1) / p + 1;
    ans = (ans + pw[p] * (phi[lim] - 1 + mod) % mod) % mod;
    ans = (ans - (phi[lim] - phi[lim - 1] + mod) % mod * pw[lim * p - n] % mod +
           mod) %
          mod;
  }
  printf("%lld\n", ans * qpow(k, mod - 1 - n) % mod);
}
}  // namespace zzc
int main() {
  zzc::work();
  return 0;
}

### 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;
namespace zzc {
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
const long long maxn = 1e5 + 5;
const long long mod = 1e9 + 7;
long long n, k, ans, cnt;
long long p[maxn], phi[maxn], pw[maxn];
bool vis[maxn];
long long qpow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
void init() {
  phi[1] = 1;
  for (register int i = 2; i <= n; i++) {
    if (!vis[i]) {
      p[++cnt] = i;
      phi[i] = i - 1;
    }
    for (register int j = 1; j <= cnt && i * p[j] <= n; j++) {
      vis[i * p[j]] = true;
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      }
      phi[i * p[j]] = phi[i] * (p[j] - 1);
    }
  }
  for (register int i = 1; i <= n; i++) phi[i] = (phi[i] + phi[i - 1]) % mod;
}
void work() {
  n = read();
  k = read();
  init();
  pw[0] = 1;
  for (register int i = 1; i <= n; i++) pw[i] = pw[i - 1] * k % mod;
  for (register long long i = 1; i <= n - 2; i++)
    ans = (ans + pw[i] * (n - 1 - i) % mod);
  pw[0] = 0;
  for (register long long p = 1; p < n; p++) {
    int lim = (n - 1) / p + 1;
    ans = (ans + pw[p] * (phi[lim] - 1 + mod) % mod) % mod;
    ans = (ans - (phi[lim] - phi[lim - 1] + mod) % mod * pw[lim * p - n] % mod +
           mod) %
          mod;
  }
  printf("%lld\n", ans * qpow(k, mod - 1 - n) % mod);
}
}  // namespace zzc
int main() {
  zzc::work();
  return 0;
}

```