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 <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

typedef long long ll;

const int N=1e5+5,Mo=1e9+7;

int pwr(int x,int y) {
	int z=1;
	for(;y;y>>=1,x=(ll)x*x%Mo)
		if (y&1) z=(ll)z*x%Mo;
	return z;
}

int n,k,a[N],tot,ans,f[N];

int main() {
	scanf("%d%d",&n,&k);
	int m=sqrt(n);
	fo(i,1,m) 
		if (!(n%i)) {
			a[++tot]=i;
			if (i*i!=n) a[++tot]=n/i;
		}
	sort(a+1,a+tot+1);
	int inv=pwr(2,Mo-2);
	fo(i,1,tot) {
		f[i]=pwr(k,(a[i]+1)/2);
		fo(j,1,i-1) if (!(a[i]%a[j])) (f[i]+=Mo-f[j])%=Mo;
		if (a[i]&1) (ans+=(ll)f[i]*a[i]%Mo)%=Mo;
		else (ans+=(ll)f[i]*a[i]%Mo*inv%Mo)%=Mo;
	}
	printf("%d\n",ans);
	return 0;
}