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: Wrong Answer)

#include <iostream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cassert>
#include <algorithm>
#include <iterator>
#include <string>
#include <vector>
#include <list>
#include <deque>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <tuple>
#include <queue>
#include <stack>
#include <functional>
#include <utility>
#include <complex>
#include <bitset>
#include <numeric>
#include <random>
using namespace std;

#define REP(i,n) for(int (i)=0; (i)<(n) ;++(i))
#define REPN(i,a,n) FOR((i),(a),(a)+(n))
#define FOR(i,a,b) for(int (i)=(a); (i)<(b) ;++(i))
#define PB push_back
#define MP make_pair
#define SE second
#define FI first
#define DBG(a) cerr<<(a)<<endl;
#define ALL(v) (v).begin(),(v).end()
typedef long long LL;  typedef pair<LL, LL> PLL; typedef vector<LL> VLL;
const LL LINF=334ll<<53; const int INF=15<<26; const LL MOD=1E9+7;

long long modpow(long long a, long long n){
    if(n==0) return 1;
    long long re=modpow((a*a)%MOD,n/2)%MOD;
    if(n%2==1) re= (re*a)%MOD;
    return re;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    LL n,k;
    cin >> n >> k;
    if(n%2==0){
        cout << 0 <<endl;
    }else{
        cout << (n*modpow(k,(n+1)/2)-(n-1)*k+MOD)%MOD<<endl;
    }


    return 0;
}