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<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
#define LL long long
using namespace std;
inline int read( )
{
  int sum=0;char c=getchar( );bool f=0;
  while(c<'0'||c>'9') {if(c=='-') f=1;c=getchar( );}
  while(c>='0'&&c<='9') {sum=sum*10+c-'0';c=getchar( );}
  if(f) return -sum;
  return sum;
}
const int mod=1000000007;
inline LL quick(LL a,LL b)
{
  LL r=1,base=a;
  while(b)
    {
      if(b&1) r=r*base%mod;
      base=base*base%mod;
      b>>=1;
    }
  return r;
}
int p[10005],f[10005];
inline void up(int &x,int y) {x+=y;if(x>=mod) x-=mod;}
int main( )
{
  int n,k,i,j,x,ans=0,len=0;
  n=read( );k=read( );
  for(i=1;i*i<=n;i++)
    if(n%i==0) p[++len]=i,p[++len]=n/i;
  sort(p+1,p+len+1);
  len=unique(p+1,p+len+1)-p-1;
  for(i=1;i<=len;i++)
    {
      x=p[i];
      f[i]=quick(k,(x+1)>>1);
      for(j=1;j<i;j++)
	if(x%p[j]==0) up(f[i],mod-f[j]);
      if(!(x&1)) x>>=1;
      up(ans,(LL)f[i]*x%mod);
    }
  printf("%d",ans);
  return 0;
}
