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>
using namespace std;
typedef long long LL;
const int N=200005,p=1000000007;
int f[N],d[N],d2[N],i,j,k,n,m,K,g,an;
char c;

inline char getc()
{
	#define VV 10000000
	static char s[VV],*l=s,*r=s;
	if (l==r)
		l=s,r=s+fread(s,1,VV,stdin);
	return l==r?0:*l++;
}

int read(){ int z=0; do c=getc(); while (c<'0'||c>'9'); while (c>='0'&&c<='9') z*=10,z+=c-'0',c=getc(); return z; }

void minus(int &a,int b){ a-=b,a+=a<0?p:0; }

int power(int z,int k)
{
	int d=1;
	for (;k;z=(LL)z*z%p,k>>=1) if (k&1) d=(LL)d*z%p;
	return d;
}

int main()
{
	n=read(),K=read();
	for (i=1;i*i<n;++i)
		if (!(n%i))
			d[++g]=i,d2[++d2[0]]=n/i;
	if (i*i==n) d[++g]=i;
	for (i=d2[0];i;--i) d[++g]=d2[i];
	for (i=1;i<=g;++i)
	{
		f[i]=power(K,d[i]+1>>1);
		for (j=1;j<i;++j)
			if (!(d[i]%d[j]))
				minus(f[i],f[j]);
		an=(an+(d[i]&1?(LL)f[i]*d[i]:(LL)f[i]*d[i]>>1))%p;
	}
	printf("%d\n",an);
	return 0;
}