#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int inf = 0x3f3f3f3f;
inline long long read() {
  register long long x = 0, f = 1;
  register char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return x * f;
}
void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
void wri(long long x) { write(x), putchar(' '); }
void writeln(long long x) { write(x), putchar('\n'); }
const int maxn = 1e5 + 233;
bool isp[maxn];
long long phi[maxn], p[maxn], ans[maxn], sum[maxn];
int n, k;
long long mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = ans * a % mod;
  return ans;
}
int main() {
  n = read(), k = read();
  memset(isp, true, sizeof(isp));
  phi[1] = 1;
  for (register int i = (2); i <= int(n); ++i) {
    if (isp[i]) {
      phi[i] = i - 1;
      p[++*p] = i;
    }
    for (int j = 1; j <= *p && i * p[j] <= n; ++j) {
      isp[i * p[j]] = false;
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      } else
        phi[i * p[j]] = phi[i] * phi[p[j]];
    }
  }
  for (register int i = (2); i <= int(n); ++i) sum[i] = sum[i - 1] + phi[i];
  for (register int i = (0); i <= int(n - 1); ++i) ans[i] = n - i - 1;
  for (register int d = (1); d <= int(n - 1); ++d) {
    int x = (n + d - 1) / d;
    ans[d] += sum[x];
    ans[x * d - n] -= phi[x];
  }
  long long tot = 0, dq = 1;
  for (register int i = (0); i <= int(n - 1); ++i) {
    tot = (tot + ans[i] * dq) % mod;
    dq = dq * k % mod;
  }
  writeln(tot * qpow(k, mod - 1 - n) % mod);
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int inf = 0x3f3f3f3f;
inline long long read() {
  register long long x = 0, f = 1;
  register char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return x * f;
}
void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
void wri(long long x) { write(x), putchar(' '); }
void writeln(long long x) { write(x), putchar('\n'); }
const int maxn = 1e5 + 233;
bool isp[maxn];
long long phi[maxn], p[maxn], ans[maxn], sum[maxn];
int n, k;
long long mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = ans * a % mod;
  return ans;
}
int main() {
  n = read(), k = read();
  memset(isp, true, sizeof(isp));
  phi[1] = 1;
  for (register int i = (2); i <= int(n); ++i) {
    if (isp[i]) {
      phi[i] = i - 1;
      p[++*p] = i;
    }
    for (int j = 1; j <= *p && i * p[j] <= n; ++j) {
      isp[i * p[j]] = false;
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      } else
        phi[i * p[j]] = phi[i] * phi[p[j]];
    }
  }
  for (register int i = (2); i <= int(n); ++i) sum[i] = sum[i - 1] + phi[i];
  for (register int i = (0); i <= int(n - 1); ++i) ans[i] = n - i - 1;
  for (register int d = (1); d <= int(n - 1); ++d) {
    int x = (n + d - 1) / d;
    ans[d] += sum[x];
    ans[x * d - n] -= phi[x];
  }
  long long tot = 0, dq = 1;
  for (register int i = (0); i <= int(n - 1); ++i) {
    tot = (tot + ans[i] * dq) % mod;
    dq = dq * k % mod;
  }
  writeln(tot * qpow(k, mod - 1 - n) % mod);
  return 0;
}

```