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 <cstring>
#include <algorithm>

using namespace std;

const int N=10000,mo=1e9+7;

typedef long long LL;

int n,m,tot,p[N],f[N],Inv,ans;

int quick(int x,int y)
{
	if (!y) return 1;
	int s=quick(x,y>>1); s=(LL)s*s%mo;
	if (y&1) s=(LL)s*x%mo;
	return s;
}

int main()
{
	scanf("%d%d",&n,&m);
	Inv=quick(2,mo-2);
	for (int i=1;i*i<=n;i++) if (n%i==0)
	{
		p[tot++]=i;
		if (i*i<n) p[tot++]=n/i;
	}
	sort(p,p+tot);
	for (int i=0;i<tot;i++)
	{
		f[i]=quick(m,(p[i]+1)/2);
		for (int j=0;j<i;j++) if (p[i]%p[j]==0) f[i]=(f[i]-f[j])%mo;
		if (f[i]<0) f[i]+=mo;
		if (p[i]&1) ans=(ans+(LL)f[i]*p[i])%mo;
		else ans=(ans+(LL)f[i]*p[i]%mo*Inv)%mo;
	}
	printf("%d\n",ans);
    return 0;
}