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 <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 1000000;
const int MOD = 1e9 + 7;
const int INV_2 = (MOD + 1) >> 1;
int n, K;
int s[MAXN], tot, f[MAXN];
int ans;

int quick_pow(int a, int x) {
    int res = 1;
    while (x) {
        if (x & 1)
            res = 1ll * res * a % MOD;
        a = 1ll * a * a % MOD, x >>= 1;
    }
    return res;
}

int main() {
	scanf("%d %d", &n, &K);
    for (int i = 1; i * i <= n; i++)
        if (!(n % i)) {
            s[++tot] = i;
            if (i * i != n)
                s[++tot] = n / i;
        }
    sort(s + 1, s + tot + 1);
    for (int i = 1; i <= tot; i++) {
        f[i] = quick_pow(K, (s[i] + 1) >> 1);
        for (int j = 1; j < i; j++)
            if (!(s[i] % s[j]))
                f[i] = (f[i] - f[j] + MOD) % MOD;
        ans = (ans + 1ll * f[i] * ((s[i] & 1) ? s[i] : s[i] >> 1)) % MOD;
    }
    printf("%d", ans);
	return 0;
}