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 <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll;
#define rep(i, a, b) for (int i = a; i <= b; ++ i)
const int mo = 1000000007;
const int maxn = 100010;
using namespace std;
int n, k, a[maxn], f[maxn];
int mul(int a, int b) {
	ll ans = 1, w = a % mo;
	for ( ; b; b /= 2, w = w * w % mo)
		if (b & 1) ans = ans * w % mo;
	return (int) ans; 
}
int main() {
	scanf("%d%d", &n, &k); a[0] = 0;
	for (int i = 1; i * i <= n; ++ i) if (n % i == 0)
		a[++ a[0]] = i, a[++ a[0]] = n / i;
	if (a[a[0]] == a[a[0] - 1]) -- a[0];
	sort(a + 1, a + a[0] + 1);
	rep(i, 1, a[0]) f[i] = mul(k, (a[i] + 1) / 2);
	rep(i, 1, a[0]) rep(j, 1, i - 1) if (a[i] % a[j] == 0)
		f[i] = (f[i] - f[j] + mo) % mo;
	int ans = 0, INV = mul(2, mo - 2);
	rep(i, 1, a[0]) if (a[i] % 2 == 0) f[i] = (ll) f[i] * INV % mo;
	rep(i, 1, a[0]) ans = (ans + (ll) f[i] * a[i] % mo) % mo;
	printf("%d", ans);
	return 0;
}