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<cmath>
#include<algorithm>
#define LL long long
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=beg[j];i;i=nex[i])
using namespace std;
int const mn=5e3+3,mo=1e9+7;
int n,K,a[mn],f[mn];
int Pow(LL x,int y){
	LL ans=1;
	for(;y;y/=2,(x*=x)%=mo)if(y&1)(ans*=x)%=mo;
	return ans;
}
int main(){
	scanf("%d%d",&n,&K);
	int lim=sqrt(n);
	fo(i,1,lim)if(n%i==0){
		a[++a[0]]=i;
		if(i*i!=n)a[++a[0]]=n/i;
	}
	sort(a+1,a+a[0]+1);
	int ans=0;
	fo(i,1,a[0]){
		f[i]=Pow(K,(a[i]+1)/2);
		fo(j,1,i-1)if(a[i]%a[j]==0)f[i]=(f[i]-f[j])%mo;
		ans=(ans+1ll*f[i]*((a[i]&1)?a[i]:a[i]/2))%mo;
	}
	printf("%d",(ans+mo)%mo);
	return 0;
}
