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 ll long long
using namespace std;
ll n,m,ans,a[5010],cnt,f[5010],p=1e9+7;
ll ksm(ll x,ll y){
	ll xlh=1;
	while(y){
		if(y%2==1)xlh=xlh*x%p;
		x=x*x%p;
		y/=2;
	}
	return xlh;
}
int main(){
	ll i,j;
	scanf("%lld%lld",&n,&m);
	for(i=1;i*i<=n;i++)if(n%i==0){
		a[++cnt]=i;
		if(n/i!=i)a[++cnt]=n/i;
	}
	sort(a+1,a+cnt+1);
	for(i=1;i<=cnt;i++){
		f[i]=ksm(m,(a[i]+1)/2)%p;
		for(j=1;j<i;j++)if(a[i]%a[j]==0)f[i]=(f[i]-f[j]+p)%p;
		if(a[i]%2==1)ans=(ans+a[i]*f[i]%p)%p;
		 else ans=(ans+a[i]*f[i]%p*ksm(2,p-2)%p)%p;
	}
	printf("%lld",ans);
}