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<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<stack>
using namespace std;
typedef long long ll;
#define N 1000015
const int p=1e9+7;
int n,k,a[N],tot,dp[N];
inline int ksm(int d,int tk){int ret=1;while(tk){if(tk&1)ret=1ll*ret*d%p;d=1ll*d*d%p;tk>>=1;}return ret;}
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i*i<=n;i++)if(n%i==0)
    {
    	a[++tot]=i;if(i*i!=n)a[++tot]=n/i;
	}
	sort(a+1,a+1+tot);int ans=0;
	for(int i=1;i<=tot;i++)
	{
		dp[i]=ksm(k,(a[i]+1)/2);
		for(int j=1;j<i;j++)if(a[i]%a[j]==0)dp[i]=(dp[i]-dp[j]+p)%p;
		if(a[i]%2==0)ans=(ans+1ll*dp[i]*(a[i]/2)%p)%p;
		else ans=(ans+1ll*dp[i]*a[i]%p)%p;
	}printf("%d\n",ans);
}
