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 double db;
typedef long long ll;
#define w1 first
#define w2 second
#define ls (x<<1)
#define rs (x<<1|1)
#define pb push_back
#define mid ((l+r)>>1)
#define SZ(x) ((x).size())
#define All(x) (x).begin(),(x).end()
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define rep2(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define per(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)
#define Rep(p,x) for(int (p)=head[(x)];(p);(p)=nxt[(p)])
template<class T>void read(T&num){
	num=0;T f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')num=num*10+ch-'0',ch=getchar();
	num*=f;
}
int power(int x,int k,int p){int res=1;for(;k;k>>=1,x=1ll*x*x%p)if(k&1)res=1ll*res*x%p;return res;}
const int maxn=1e6+5,mod=1e9+7;
int n,k,ans;
int sub(int x,int y){return x-y<0?x-y+mod:x-y;}
vector<int>a;
int f[maxn];
int main(){
	read(n);read(k);
	int tmp=sqrt(n);
	rep(i,1,tmp)if(n%i==0){
		a.pb(i);
		if(n/i!=i)a.pb(n/i);
	}
	sort(All(a));
	ans=f[0]=k;
	rep2(i,1,SZ(a)){
		f[i]=power(k,(a[i]+1)/2,mod);
		rep2(j,0,i)if(a[i]%a[j]==0)f[i]=sub(f[i],f[j]);
		if(a[i]&1)ans=(ans+1ll*f[i]*a[i])%mod;
		else ans=(ans+1ll*f[i]*(a[i]/2))%mod;
	}
	printf("%d\n",ans);
	return 0;
}