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 mod = 1e9 + 7;
vector<int> d;
int n, k, cnt = -1;
long long dp[12911], ans;
map<int, int> id;

inline long long fpow(int a, int b){
	long long ret = 1, tmp = a;
	while(b){
		if(b & 1) ret = ret * tmp % mod;
		tmp = tmp * tmp % mod; b >>= 1;
	}
	return ret;
}

int main()
{
	scanf("%d%d", &n, &k); int sq = sqrt(n);
	if(n == 1){printf("%d", k); return 0;}
	for(int i = 1; i <= sq; ++i) 
		if(n % i == 0){d.push_back(i); id[i] = ++cnt;}
	for(int i = d.size() - 1; i >= 0; --i) 
		if(n / d[i] > sq){d.push_back(n / d[i]); id[n / d[i]] = ++cnt;}
	int llppdd = d.size() - 1;
	for(int i = 0; i <= llppdd; ++i){
		int now = d[i];
		if((n / now) & 1) 
		dp[i] = fpow(k, (now + 1) / 2);
		else 
		dp[i] = fpow(k, now);
		sq = sqrt(now);
		for(int j = 1; j <= sq; ++j){
			if(now % j != 0 || j == now) continue;
			dp[i] = (dp[i] - dp[id[j]] + mod) % mod;
			if(now / j > sq && j != 1) dp[i] = (dp[i] - dp[id[now / j]] + mod) % mod;
		}
		ans = (ans + dp[i] * now) % mod;
	}
	printf("%lld", ans);
	return 0;
}