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;
typedef long long LL;
const int N=2000,mod=1e9+7;
inline int powmod(int a,int b){int c=1;while(b){if(b&1)c=(LL)c*a%mod;a=(LL)a*a%mod;b>>=1;}return c;}
int n,k,f[N],g[N],num,ans;
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i*i<=n;i++)if(n%i==0)
	{
		f[++num]=i;
		if(i*i!=n)f[++num]=n/i;
	}
	sort(f+1,f+num+1);
	for(int i=1;i<=num;i++)
	{
		g[i]=powmod(k,(f[i]+1)/2);
		for(int j=1;j<i;j++)if(f[i]%f[j]==0)g[i]=(g[i]-g[j]+mod)%mod;
		ans=(ans+(LL)g[i]*((f[i]&1)?f[i]:f[i]/2)%mod)%mod;
	}
	printf("%d\n",ans);
	return 0;
}