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<algorithm>
#include<iostream>
#include<cstdio>
#define mod 1000000007
using namespace std;
int n,K,p[105],c[105],cnt,a[5005],top,f[5005],ans=0;
void dfs(int s,int x)
{
	if(s>cnt){a[++top]=x;return;}
	for(int k=0;k<=c[s];++k) dfs(s+1,x),x*=p[s];
}
inline int pow(int x,int k)
{
	int sum=1;
	for(;k;k>>=1,x=1LL*x*x%mod) if(k&1) sum=1LL*sum*x%mod;
	return sum;	
}
int main()
{
	scanf("%d%d",&n,&K);int t=n;
	for(int i=2;i*i<=t;++i) if(t%i==0)
	{
		p[++cnt]=i;
		while(t%i==0) t/=i,++c[cnt];
	}
	if(t>1) p[++cnt]=t,c[cnt]=1;
	dfs(1,1);sort(a+1,a+top+1);
	for(int i=1;i<=top;++i)
	{
		f[i]=pow(K,(a[i]+1)>>1);
		for(int j=1;j<i;++j) if(a[i]%a[j]==0) f[i]=(f[i]-f[j]+mod)%mod;
		ans=(ans+1LL*f[i]*((a[i]&1)?a[i]:a[i]/2))%mod;
	}
	cout<<ans;
	return 0;
}