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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#define mod 1000000007
#define ll long long
using namespace std;
int n,k,tot;
int s[2005],f[2005];
int po(int a,int b){
	int an=1;
	while(b!=0){
		if(b&1) an=(ll)an*a%mod;
		a=(ll)a*a%mod;b=(b>>1);
	}
	return an;
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i*i<=n;i++)
		if(n%i==0){
			s[++tot]=i;
			if(i*i!=n) s[++tot]=n/i; 
		}
	sort(s+1,s+1+tot);f[1]=k;
	for(int i=2;i<=tot;i++){
		f[i]=po(k,(s[i]+1)/2);
		for(int j=1;j<i;j++)
			if(s[i]%s[j]==0) f[i]=(f[i]-f[j]+mod)%mod;
	}
	int ans=0,ni=po(2,mod-2);
	for(int i=1;i<=tot;i++){
		if(s[i]%2==0) ans=(ans+(ll)f[i]*ni%mod*s[i]%mod)%mod;
		else ans=(ans+(ll)f[i]*s[i]%mod)%mod; 
	}
	printf("%d\n",ans);
	return 0;
}