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>
#define fi first
#define se second
#define ll long long
using namespace std;

const ll MOD=1e9+7;
ll fpow(ll a,ll p){
	ll res=1LL;
	while(p){
		if(p&1LL) res=res*a%MOD;
		a=a*a%MOD;
		p>>=1LL;
	}
	return res;
}
ll n,k;
vector<ll> divisors,dp;

int main() {
	#ifdef kvutxdy
	freopen("C:/Users/khanh/OneDrive/RoadtoPurple/Code/Codeforces/input.in", "r", stdin);
	#endif
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	cin>>n>>k;
	for(ll i=1;i*i<=n;++i) if(n%i==0){
		divisors.push_back(i);
		if(i*i!=n) divisors.push_back(n/i);
	}
	sort(divisors.begin(),divisors.end());
	int m=divisors.size();
	ll res=0LL;
	dp.resize(m);
	for(int i=0;i<m;++i){
		dp[i]=fpow(k,(divisors[i]+1)/2);
		for(int j=0;j<i;++j) if(divisors[i]%divisors[j]==0){
			dp[i]-=dp[j];
			if(dp[i]<0) dp[i]+=MOD;
		}
		ll tmp=dp[i]*divisors[i]%MOD;
		if(divisors[i]%2==0) (tmp*=fpow(2LL,MOD-2LL))%=MOD;
		(res+=tmp)%=MOD; 
	}
	cout<<res;
}