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 mo 1000000007
using namespace std;
int n,k,ans,divi[2333];
map<int,int> mp;
int power(int x,int y){
	int s=1;
	for (;y;y/=2,x=1ll*x*x%mo)
		if (y&1) s=1ll*s*x%mo;
	return s;
}
int calc(int x){
	if(mp.find(x)!=mp.end()) return mp[x];
	int ans=power(k,(x+1)/2);
	for (int i=1;i<=*divi&&divi[i]<x;i++)
		if (x%divi[i]==0) ans=(ans+mo-calc(divi[i]))%mo;
	//printf("%d %d\n",x,ans);
	return mp[x]=ans;
}
int main(){
	scanf("%d%d",&n,&k);
	for (int i=1;i*i<=n;i++)
		if (n%i==0){
			divi[++*divi]=i;
			if (i*i!=n) divi[++*divi]=n/i;
		}
	sort(divi+1,divi+*divi+1);
	for (int i=1;i<=*divi;i++)
		ans=(ans+1ll*calc(divi[i])*(divi[i]/(2-(divi[i]&1))))%mo;
	printf("%d",ans);
}