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;

#define MOD 1000000007

int n;
int k;


vector<int> dv;
vector<long long int>  way;

long long int ppow(long long int i, long long int j) {
	long long int res = 1LL;
	while (j) {
		if ((j & 1LL)) {
			res *= i;
			if (res >= MOD) {
				res %= MOD;
			}
		}
		j >>= 1;
		i *= i;
		if (i >= MOD) {
			i %= MOD;
		}
	}
	return res;
}

int main() {
	cin >> n >> k;
	for (long long int i = 1; i*i <= n; i++) {
		if(n%i==0)dv.push_back(i);
		if (n%i == 0)dv.push_back(n / i);
	}
	sort(dv.begin(), dv.end());
	dv.erase(unique(dv.begin(), dv.end()), dv.end());
	for (int i = 0; i < dv.size(); i++) {
		way.push_back(ppow(k, (dv[i] + 1) / 2));
		for (int j = 0; j < i; j++){
			if (dv[i] % dv[j] == 0) {
				way[i] += MOD - way[j];
				if (way[i] >= MOD)way[i] %= MOD;
			}
		}
	}
	long long int ans = 0;
	for (int i = 0; i < dv.size(); i++) {
		if (dv[i] % 2 == 0) {
			ans += way[i] * dv[i]/2;
		}
		else {
			ans += way[i] * dv[i];
		}
		if (ans >= MOD)ans %= MOD;
	}
	printf("%lld\n", ans);
	return 0;
}