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 <vector>
#include <algorithm>
const int mo = 1e9 + 7;
using namespace std;
int pow(int x, int t){
	int res = 1;
	while (t > 0){
		if (t & 1) res = 1ll * res * x % mo;
		x = 1ll * x * x % mo;
		t >>= 1;
	}
	return res;
}
int n, k, dp[10001]; vector <int> p;
int main(){
	scanf("%d%d", &n, &k);
	for (int i = 1; i * i <= n; i++){
		if (n % i == 0){
			p.push_back(i);
			if (i * i != n) p.push_back(n / i);
		}
	}
	sort(p.begin(), p.end()); int ans = 0;
	for (int i = 0; i < p.size(); i++){
		dp[i] = pow(k, (p[i] + 1) >> 1);
		for (int j = 0; j < i; j++)
			if (p[i] % p[j] == 0) dp[i] = (dp[i] - dp[j] + mo) % mo;
		ans = (ans + 1ll * dp[i] * (p[i] / (1 + (p[i] % 2 == 0))) % mo) % mo;
	}
	printf("%d\n", ans);
	return 0;
}