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>
#define fo(i,a,b) for(i=a;i<=b;i++)
#define ll long long

using namespace std;

const int maxn=1e5+5,md=1e9+7;

ll a[maxn],f[maxn];
ll i,j,k,l,m,n,x,y,ans,now,p;

ll quickmi(ll x,ll y){
	ll t1=1;
	while (y){
		if ((y & 1)) t1=(t1*x)%md;
		x=(x*x)%md;
		y/=2; 
	}
	return t1;
}
int main(){
	while (cin>>n>>k){
	p=n; now=0;
	for(i=1;i*i<=p;i++){
		if (p%i==0){
			a[++now]=i;
			if ((p/i)!=i) a[++now]=p/i;
		}
	}
	sort(a+1,a+now+1);
	memset(f,0,sizeof(f)); ans=0;
	fo(i,1,now){
		f[i]=quickmi(k,(a[i]+1)/2);
		fo(j,1,i-1) if (a[i]%a[j]==0) f[i]=(f[i]-f[j]+md)%md;
		if (a[i]%2==0) ans=(ans+f[i]*(a[i]/2))%md;
		else ans=(ans+f[i]*a[i])%md;
	}
	printf("%d\n",ans);
	}
	return 0;
}