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;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> ii;
typedef vector<int> vi;

#define SZ(x) ((int)x.size())

const ll base = (ll)1e9+7;
ll n, k, dp[5005];
vector<ll> d;

ll fast_pw(ll x, ll y) {
	if (y == 0) return 1;
	if (y == 1) return x;
	ll tmp = fast_pw(x, y / 2ll);
	tmp = (tmp * tmp) % base;
	if (y % 2ll) tmp = (tmp * x) % base;
	return tmp;
}

int main() {
	scanf("%lld%lld", &n, &k);
	for (ll i = 1; i*i <= n; ++i) if (n % i == 0) {
		d.push_back(i);
		if (n / i != i) d.push_back(n / i);
	}
	sort(d.begin(), d.end());
	ll ans = 0;
	for (int i = 0; i < SZ(d); ++i) {
		dp[i] = fast_pw(k, d[i] / 2 + d[i] % 2);
		for (int j = 0; j < i; ++j)
			if (d[i] % d[j] == 0)
				dp[i] = (dp[i] - dp[j] + base) % base;
		if (d[i] % 2 == 0) ans = (ans + dp[i] * (ll)(d[i] / 2) % base) % base;
		else ans = (ans + dp[i] * (ll)d[i] % base) % base;
	}
	printf("%lld", ans);
	return 0;
}