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 N 11116
#define mod 1000000007

int n,k;
vector<int> q;
long long f[N],ans;

long long power(long long x,long long y){
	long long t=1;
	for (;y;x=x*x%mod,y/=2)
		if (y&1) t=t*x%mod;
	return t;
}

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read();k=read();
	for (int i=1;i*i<=n;i++)
		if (n%i==0){
			q.push_back(i);
			if (i*i!=n) q.push_back(n/i);
		}
	sort(q.begin(),q.end());
	for (int i=0;i<(int)q.size();i++){
		f[i]=power(k,(q[i]+1)/2);
		for (int j=0;j<i;j++)
			if (q[i]%q[j]==0)
			f[i]=(f[i]-f[j])%mod;
		if (q[i]%2)
		ans=(ans+f[i]*q[i])%mod;
		else ans=(ans+f[i]*q[i]/2)%mod;
	}
	printf("%lld\n",(ans+mod)%mod);
	return 0;
}