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>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define N 10100
#define mo 1000000007
#define ll long long
using namespace std;
int n,m,k,f[N],ans=0,a[N];
int mi(ll a,ll b)
{
	ll c=1;
	for(;b;b/=2,a=a*a%mo) if(b%2==1) c=c*a%mo;
	return c;
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i*i<=n;i++)
	if(n%i==0)
	{
		a[++m]=i;
		if(i!=n/i) a[++m]=n/i;
	}
	sort(a+1,a+m+1);
	fo(i,1,m)
	{
		f[i]=mi(k,(a[i]+1)/2);
		fo(j,1,i-1)
		{
			if(a[i]%a[j]==0) f[i]=(f[i]-f[j]+mo)%mo;
		}
		ans=(ans+(ll)f[i]*(a[i]%2==1?(ll)a[i]:(ll)(a[i]+1)/2)%mo)%mo;
	}
	printf("%d\n",ans);
}