Takahashi and Aoki are going to together construct a sequence of integers.

First, Takahashi will provide a sequence of integers a, satisfying all of the following conditions:

* The length of a is N.
* Each element in a is an integer between 1 and K, inclusive.
* a is a palindrome, that is, reversing the order of elements in a will result in the same sequence as the original.



Then, Aoki will perform the following operation an arbitrary number of times:

* Move the first element in a to the end of a.



How many sequences a can be obtained after this procedure, modulo 10^9+7?

Constraints

* 1≤N≤10^9
* 1≤K≤10^9

Input

The input is given from Standard Input in the following format:


N K


Output

Print the number of the sequences a that can be obtained after the procedure, modulo 10^9+7.

Examples

Input

4 2


Output

6


Input

1 10


Output

10


Input

6 3


Output

75


Input

1000000000 1000000000


Output

875699961
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

typedef long long i64;
const int MOD = 1000000007;

int N, K, tot;
int divisor[100005];
i64 f[100005];

i64 fpm(i64 x, i64 y) {
  i64 res = 1;
  while (y) {
    if (y & 1) res = res * x % MOD;
    x = x * x % MOD;
    y >>= 1;
  }
  return res;
}

int main() {
  scanf("%d%d", &N, &K);
  
  for (int i = 1; i * i <= N; ++i)
    if (N % i == 0) {
      divisor[++tot] = i;
      if (N / i != i) divisor[++tot] = N / i;
    }

  sort(divisor + 1, divisor + tot + 1);
  
  for (int i = 1; i <= tot; ++i) {
    f[i] = fpm(K, (divisor[i] + 1) / 2);
    for (int j = 1; j < i; ++j)
      if (divisor[i] % divisor[j] == 0)
	f[i] = (f[i] - f[j]) % MOD;
  }

  i64 res = 0;

  for (int i = 1; i <= tot; ++i)
    if (divisor[i] & 1)
      res = (res + f[i] * divisor[i] % MOD) % MOD;
    else
      res = (res + f[i] * divisor[i] / 2 % MOD) % MOD;

  printf("%lld\n", (res + MOD) % MOD);
  return 0;
}
