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 <bits/stdc++.h>
using namespace std;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1100000000000000000

int beki(int a,int b,int M = modulo){
	int x = 1;
	while(b!=0){
		if(b&1){
			x=((long long)x*a)%M;
		}
		a=((long long)a*a)%M;
		b>>=1;
	}
	return x;
}


int gyakugen(int a){
	return beki(a,modulo-2);
}

int main(){
	
	int N,K;
	cin>>N>>K;
	
	vector<int> Y;
	for(int i=1;i*i<=N;i++){
		if(N%i==0){
			Y.push_back(i);
			if(i*i!=N)Y.push_back(N/i);
		}
	}
	
	sort(Y.begin(),Y.end());
	vector<int> c(Y.size(),0);
	
	for(int i=0;i<Y.size();i++){
		c[i] = beki(K,(Y[i]+1)/2);
		
		for(int j=0;j<i;j++){
			if(Y[i]%Y[j]==0){
				int x = c[j];
				c[i] = mod(c[i]-x);
			}
		}
	}
	
	int ans = 0;
	
	for(int i=0;i<Y.size();i++){
		if(Y[i]%2==0)ans = mod(ans + mod(Y[i] * c[i]/2));
		else ans = mod(ans + mod(Y[i] * c[i]));
	}
	
	cout<<ans<<endl;
	
	return 0;
}
