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 <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <bitset>
#include <string>
#include <cmath>
#include <set>
#include <map>
#define Rep(i, x, y) for (int i = x; i <= y; i ++)
#define Dwn(i, x, y) for (int i = x; i >= y; i --)
#define RepE(i, x) for (int i = pos[x]; i; i = g[i].nex)
#define ev g[i].y
#define eps 1e-10
using namespace std;
typedef long long ll;
typedef double db;
const int N = 10005, mod = 1000000007;
int n, m, k, a[N];
ll f[N], ans;
// struct arr { int x; };
ll mypow(ll x, int y) {
	ll z = 1;
	while (y) {
		if (y & 1) (z *= x) %= mod;
		y >>= 1; (x *= x) %= mod;
	}
	return z;
}
int main()
{
	scanf ("%d%d", &n, &k);
	for (int i = 1; i * i <= n; i ++) {
		if (n % i == 0) {
			a[++ m] = i;
			if (i * i != n) {
				a[++ m] = n / i;
			}
		}
	}
	sort(a + 1, a + m + 1);
	Rep(i, 1, m) {
		int x = a[i];
		f[i] = mypow(k, (x + 1) / 2);
		Rep(j, 1, i - 1) if (x % a[j] == 0) f[i] = (f[i] - f[j] + mod) % mod;
		// cout << f[i]<<endl;
		if (x & 1) ans += f[i] * x;
		else ans += f[i] * x / 2;
	}
	printf("%lld\n", ans % mod);

	return 0;
}