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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
ll quick_pow(ll a,ll b){
	ll ans=1;
	for(;b;b>>=1,a=a*a%mod)
		if(b&1) ans=ans*a%mod;
	return ans;
}
ll n,K,ans=0,op[100005],s[100005],sum[100005];
int top;
int main(){
	scanf("%lld%lld",&n,&K);
	for(ll i=1;i*i<=n;i++){
		if(n%i==0){
			s[++top]=i;
			if(n!=i*i) s[++top]=n/i;
		}
	}
	sort(s+1,s+top+1);
	for(int i=1;i<=top;i++) sum[i]=quick_pow(K,(s[i]+1)/2);
	for(int i=1;i<=top;i++){
		ll tot=sum[i];
		for(int p=1;p<i;p++) if(s[i]%s[p]==0){
			tot=(tot-sum[p])%mod;
		}
		sum[i]=tot;
		ans=(ans+tot*((s[i]&1)?s[i]:(s[i]>>1)))%mod;
	}
	ans=(ans+mod)%mod;
	printf("%lld",ans);
	return 0;
}