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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <vector>

using namespace std;
typedef long long LL;

const int maxn = 10005;
const int mod = 1e9+7;

int f[maxn],num[maxn],tot,n,K,ans;
int fpm(LL p,LL k) {
	LL res=1ll;
	while (k) {
		if (k&1) (res*=p)%=mod;
		(p*=p)%=mod;k>>=1;
	}
	return res;
}
int calc(int n) {
	return n&1?n:n>>1;
}

int main()
{
	#ifdef Amberframe
		freopen("arc064f.in","r",stdin);
		freopen("arc064f.out","w",stdout);
	#endif
	scanf("%d %d",&n,&K);
	for (int i=1;i*i<=n;i++)
	if (n%i==0) {
		num[++tot]=n/i;
		if (i*i!=n) num[++tot]=i;
	}
	sort(num+1,num+tot+1);
	for (int i=1;i<=tot;i++)
	{
		f[i]=fpm(K,num[i]+1>>1);
		for (int j=1;j<i;j++)
		if (num[i]%num[j]==0)
			f[i]=(f[i]-f[j])%mod;
		ans=(ans+(LL)f[i]*calc(num[i]))%mod;
	}
	printf("%d",ans<0?ans+mod:ans);
	return 0;
}