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 <algorithm>
#include <cstdio>
using namespace std;
const int N = 1e6 + 1;
const int mod = 1e9 + 7;
int n, k;
int a[N], f[N];
int add(int x, int y) {
    x+= y;
    return x >= mod ?x - mod : x;
}
int sub(int x, int y) {
    x -= y;
    return x < 0 ? x +mod : x;
}
int mul(int x, int y){
    return 1ll * x * y % mod;
}
int calc(int x, int y) {
    int z = 1;
    while (y) {
        if (y & 1) z = mul(z, x);
        x = mul(x, x), y >>= 1;
    }
    return z;
}
int main() {
    scanf("%d %d", &n, &k);
    for (int i = 1; 1ll * i * i <= n; i ++)
        if (n % i == 0) {
            a[++a[0]] = i;
            if (1ll * i * i != n)
                a[++a[0]] = n / i;
        }
    sort(a+1, a+ a[0]+ 1);
    int ans = 0;
    for (int i = 1; i <= a[0]; i ++) {
        f[i] = calc(k, ((a[i] + 1) / 2));
        for (int j = 1; j < i; j ++)
            if (a[i] % a[j] == 0)
                f[i] = sub(f[i], f[j]);
        if (a[i] & 1)
            ans = add(ans, mul(f[i], a[i]) );
        else
            ans = add(ans, mul(f[i], a[i] / 2) );
    }
    printf("%d\n", ans);
    return 0;
}