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>
#define P 1000000007
#define ll long long
using namespace std;
int n,k,f[100005],a[100005],ly;
int pw(int x,int y)
{
	int ans=1;
	for (;y;y>>=1,x=(ll)x*x%P)
		if (y&1) ans=(ll)ans*x%P;
	return ans;
}
int main()
{
	scanf("%d%d",&n,&k);
	int Ans=0;
	for (int i=1;i*i<=n;i++)
		if (n%i==0)
		{
			a[++ly]=i;
			if (i*i!=n) a[++ly]=n/i;
		}
	sort(a+1,a+ly+1);
	for (int i=1;i<=ly;i++)
	{
		f[i]=pw(k,(a[i]+1)/2);
		for (int j=1;j<i;j++)
			if (a[i]%a[j]==0) f[i]=(f[i]+P-f[j])%P;
		if (a[i]&1) Ans=(Ans+(ll)f[i]*a[i]%P)%P;
		else Ans=(Ans+(ll)f[i]*(a[i]/2)%P)%P;
	}
	/*
	for (int i=1;i<=n;i++)
	{
		dp[i]=pw(k,(i+1)/2);
		for (int j=1;j<i;j++)
			if (i%j==0) dp[i]=(dp[i]+P-dp[j])%P;
		//cout<<i<<' '<<dp[i]<<' '<<n/i<<endl;
		if (n%i==0)
			if (i&1) Ans=(Ans+(ll)dp[i]*i%P)%P;
			else Ans=(Ans+(ll)dp[i]*(i/2)%P)%P;
	}*/
	printf("%d\n",Ans);
}