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<string>
#include<math.h>
#include<stdlib.h>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<ctime>
#include<complex>
#include<list>
#include<limits.h>
#include<iomanip>
using namespace std;

const int MOD=1e9+7;
int n,m,ans;
int d[1505];
vector<int> v;

int Pow(int x,int y){
	int ret=1,mt=x;
	while(y){
		if(y&1) ret=1ll*ret*mt%MOD;
		mt=1ll*mt*mt%MOD;
		y>>=1;
	}
	return ret;
}

void add(int &x,int y){
	x+=y;
	if(x>=MOD) x-=MOD;
	if(x<0) x+=MOD;
}

int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i*i<=n;i++) if(n%i==0){
		v.push_back(i);
		if(i!=n/i) v.push_back(n/i);
	}
	if(n==1){
		printf("%d\n",m);
		return 0;
	}
	sort(v.begin(),v.end());
	for(int i=0;i<v.size();i++){
		int x=v[i];
		d[i]=Pow(m,(x+1)/2);
		for(int j=0;j<i;j++) if(x%v[j]==0) add(d[i],-d[j]);
		if(x%2==0) add(ans,1ll*d[i]*x/2%MOD);
		else add(ans,1ll*d[i]*x%MOD);
	}
	printf("%d\n",ans);
	return 0;
}