Problem:
Score : 1000 points

Problem Statement

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.

Sample Input 1

4 2

Sample Output 1

6

The following six sequences can be obtained:

(1, 1, 1, 1)

(1, 1, 2, 2)

(1, 2, 2, 1)

(2, 2, 1, 1)

(2, 1, 1, 2)

(2, 2, 2, 2)

Sample Input 2

1 10

Sample Output 2

10

Sample Input 3

6 3

Sample Output 3

75

Sample Input 4

1000000000 1000000000

Sample Output 4

875699961

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
const int N=2e5+20;
vector<ll> v;
ll n,k,dp[2500];
ll powmod(ll x,ll n)
{
	ll s=1;
	while(n)
	{
		if(n&1)
			s=(s*x)%mod;
		n>>=1;
		x=(x*x)%mod;
	}
	return s;
}
void init()
{
	v.clear();
	for(ll i=1;i*i<=n;i++)
	{
		if(n%i==0)
		{
			v.push_back(i);
			if(i*i!=n)
				v.push_back(n/i);
		}
	}
	sort(v.begin(),v.end());
}
int main()
{
	while(cin>>n>>k)
	{
		init();
		ll ans=0;
		for(int x=0;x<v.size();x++)
		{
			dp[x]=powmod(k,(v[x]+1)/2);
			for(int y=0;y<x;y++)
			{
				if(v[x]%v[y]==0)
					dp[x]=(dp[x]-dp[y]+mod)%mod;	
			}	
			if(v[x]%2)
				ans=(ans+(dp[x]*v[x])%mod)%mod;
			else
				ans=(ans+(dp[x]*v[x]/2)%mod)%mod;
		}
		cout<<ans<<endl;
	}
	return 0;
}