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<bits/stdc++.h>

using namespace std;

int ans, nr, N, K, dp[10009], dv[10009], mod = 1e9 + 7;

int add (int x, int y) {int ans = x + y; if (ans >= mod) ans -= mod; return ans;}
int substract (int x, int y) {if (x >= y) return x - y; return x - y + mod;}
int mul (int x, int y) {return 1LL * x * y % mod;}

int pow (int a, int b)
{
    int p = 1;
    for (int i=0; (1<<i) <= b; i++)
    {
        if (b & (1 << i)) p = mul (p, a);
        a = mul (a, a);
    }
    return p;
}

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

scanf ("%d %d", &N, &K);
for (int i=1; i * i<=N; i++)
    if (N % i == 0)
        dv[++nr] = i;
for (int i=nr; i>=1; i--)
    if (dv[i] * dv[i] != N)
        dv[++nr] = N / dv[i];
for (int i=1; i<=nr; i++)
{
    dp[i] = pow (K, (dv[i] + 1) / 2);
    for (int j=1; j<i; j++)
        if (dv[i] % dv[j] == 0)
            dp[i] = substract (dp[i], dp[j]);
    int coef = dv[i];
    if (coef % 2 == 0) coef /= 2;
    ans = add (ans, mul (dp[i], coef));
}
printf ("%d\n", ans);
return 0;
}
