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<cstdio>
#define ll long long
const int N=2005,mo=1e9+7;
int n,m,i,j,na,a[N],f[N],ans;
int Pow(int x,int n){
	int k=1;
	for (;n;n>>=1,x=(ll)x*x%mo)
		if (n&1) k=(ll)k*x%mo;
	return k;
}
int main(){
	scanf("%d%d",&n,&m);
	for (i=1;i*i<=n;i++) if (n%i==0) a[++na]=i;
	for (i=na;i;i--) if (a[i]*a[i]!=n) a[++na]=n/a[i];
	for (i=1;i<=na;i++){
		f[i]=Pow(m,a[i]+1>>1);
		for (j=1;j<i;j++)
			if (a[i]%a[j]==0) f[i]=(f[i]-f[j])%mo;
		ans=(ans+(ll)f[i]*(a[i]&1?a[i]:a[i]>>1))%mo;
	}
	printf("%d",(ans+mo)%mo);
}