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 D=1350,P=1000000007;
int n,K,fac[D],tot,f[D];

int power(int a,int x)
{
	int ans=1;
	for(;x;x>>=1,a=1ll*a*a%P)
		if(x&1)ans=1ll*ans*a%P;
	return ans;
}

int main()
{
	scanf("%d%d",&n,&K);
	int t=n;
	for(int i=1;i*i<=t;i++)
		if(t%i==0)
		{
			fac[++tot]=i;
			if(i*i!=t)fac[++tot]=t/i;
		}
	sort(fac+1,fac+1+tot);
	for(int i=1;i<=tot;i++)
	{
		f[i]=power(K,fac[i]+1>>1);
		for(int j=1;j<i;j++)
			if(fac[i]%fac[j]==0)
				f[i]=(f[i]-f[j])%P;
	}
	const int inv2=P+1>>1;
	int ans=0;
	for(int i=1;i<=tot;i++)
		if(fac[i]&1)ans=(ans+1ll*f[i]*fac[i])%P;
		else ans=(ans+1ll*inv2*f[i]%P*fac[i])%P;
	printf("%d\n",(ans+P)%P);
	return 0;
}
