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>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#define SF scanf
#define PF printf
#define MAXN 2010
#define MOD 1000000007
using namespace std;
typedef long long ll;
vector<int> fac;
int n;
ll k,sum[MAXN];
void get_fac(int x){
	for(int i=1;i*i<=x;i++)
		if(x%i==0){
			fac.push_back(i);
			if(i*i!=x)
				fac.push_back(x/i);
		}
	sort(fac.begin(),fac.end());
}
ll fsp(ll x,int y){
	ll res=1;
	while(y){
		if(y&1)
			res=res*x%MOD;
		x=x*x%MOD;
		y>>=1;	
	}
	return res;
}
int main(){
	SF("%d%lld",&n,&k);
	get_fac(n);
	ll ans=0;
	for(int i=0;i<fac.size();i++){
		sum[i]=fsp(k,(fac[i]+1)/2);
		for(int j=0;j<i;j++){
			if(fac[i]%fac[j]==0){
				sum[i]-=sum[j];
				sum[i]%=MOD;
			}
		}
		if(fac[i]%2)
			(ans+=sum[i]*fac[i]%MOD)%=MOD;
		else
			(ans+=sum[i]*(fac[i]/2ll)%MOD)%=MOD;
	}
	PF("%lld",(ans+MOD)%MOD);
}