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 <bits/stdc++.h>

using namespace std;

const int MOD = 1000000007;

using ll = long long;

int add(int x, int y) { return ((x += y) >= MOD) ? x - MOD : x; }
int sub(int x, int y) { return add(x, MOD - y); }
int mult(int x, int y) { return ll(x) * y % MOD; }
int pw(int x, int y) {
   int ans = 1;
   for (; y > 0; y >>= 1, x = mult(x, x)) if (y & 1) ans = mult(ans, x);
   return ans;
}

int main() {
   ios_base::sync_with_stdio(false);
   int N, K;
   cin >> N >> K;
   vector<int> divs;
   for (int i = 1; i * i <= N; ++i) {
      if (N % i == 0) {
         divs.emplace_back(i);
         if (i * i < N) divs.emplace_back(N / i);
      }
   }
   sort(divs.begin(), divs.end());
   int M = divs.size();
   vector<int> dp(M);
   int ans = 0;
   for (int i = 0; i < M; ++i) {
      int z = divs[i];
      dp[i] = pw(K, (z + 1) / 2);
      for (int j = 0; j < i; ++j) {
         if (divs[i] % divs[j] == 0) {
            dp[i] = sub(dp[i], dp[j]);
         }
      }
      ans = add(ans, mult(dp[i], (z % 2) ? z : z / 2));
   }
   cout << ans << "\n";
   return 0;
}
