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<vector>
#include<string>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<queue>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int MOD = 1e9+7;

vector<int> getDivs(int N) {
	vector<int> ret;
	for (int i = 1; i*i <= N; i++) {
		if (N%i == 0) {
			ret.push_back(i);
			if (i*i < N) ret.push_back(N/i);
		}
	}
	sort(ret.begin(), ret.end());
	return ret;
}

ll memo[2000];
ll powMod(ll x, ll p) {
	if (x == 0) return 0;
	if (p == 0) return 1;
	if (p == 1) return x;
	if (p%2 == 1) return x*powMod(x, p-1) % MOD;
	ll tmp = powMod(x, p/2);
	return tmp * tmp % MOD;
}

int main() {
	int N, K;
	cin >> N >> K;
	auto divs = getDivs(N);
	int M = divs.size();
	ll ans = 0;
	for (int i = 0; i < M; i++) {
		ll& res = memo[i];
		res = powMod(K, (divs[i]+1)/2);
		for (int j = 0; j < i; j++)
			if (divs[i]%divs[j] == 0) res -= memo[j];
		res = ((res%MOD)+MOD) % MOD;
		if (divs[i]%2 == 0) (ans += res*divs[i]/2%MOD) %= MOD;
		else (ans += res*divs[i]%MOD) %= MOD;
	}
	cout << ans << endl;
	return 0;
}
