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;

const int N = 100005;
const int md = 1e9 + 7;
int n, k;
vector<int> dvs;
int dp[N];

int pw(int x, int y) {
	int res = 1;
	while(y) {
		if (y & 1) res = 1LL * res * x % md; x = 1LL * x * x % md;
		y >>= 1;
	}
	return res;
}

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n >> k;

	// find divisors of n
	for (int i = 1; i * i <= n; ++i) {
		if (n % i == 0) {
			dvs.push_back(i);
			if (i != n / i) dvs.push_back(n / i);
		}
	}
	sort(dvs.begin(), dvs.end());

	// calculate DPs
	dp[0] = k;
	for (int i = 1; i < dvs.size(); ++i) {
		int d = dvs[i]; // minimum period
		dp[i] = pw(k, (d + d % 2) / 2);
		for (int j = 0; j < i; ++j) if (d % dvs[j] == 0) {
			dp[i] -= dp[j];
			if (dp[i] < 0) dp[i] += md;
		}
	}

	// calculate the answer
	int ans = 0;
	for (int i = 0; i < dvs.size(); ++i) {
		int d = dvs[i]; 
		ans = (ans + 1LL * dp[i] * ((d & 1) ? d : d / 2)) % md;
	}
	cout << ans << endl;
}