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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>

using namespace std;

const int MAXN = 50000 + 100;
const int MOD = 1000000000 + 7;

int n, m, k;
int d[MAXN], f[MAXN];

int fpow(int x, int p)
{
	int ret = 1;
	for(; p; p >>= 1, x = (long long)x * x % MOD)
		if(p & 1)
			ret = (long long)ret * x % MOD;
	return ret;
}

int main()
{
	ios::sync_with_stdio(false);
	// freopen("1.in", "r", stdin);
	// freopen("1.out", "w", stdout);

	cin >> n >> k;
	m = 0;
	for(int i = 1; i * i <= n; i++)
		if(n % i == 0)
		{
			d[++m] = i;
			if(i != n / i)
				d[++m] = n / i;
		}
	sort(d + 1, d + m + 1);
	for(int i = 1; i <= m; i++)
	{
		f[i] = fpow(k, (d[i] + 1) / 2);
		for(int j = 1; j <= i - 1; j++)
			if(d[i] % d[j] == 0)
				f[i] = (f[i] - f[j] + MOD) % MOD;
		// cerr << d[i] << ' ' << f[i] << endl;
	}
	int ans = 0;
	for(int i = 1; i <= m; i++)
		ans = (ans + (long long)d[i] * f[i] / (d[i] & 1 ? 1 : 2)) % MOD;
	cout << ans << endl;

	return 0;
}
