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<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#define ll long long
#define mod 1000000007
using namespace std;
int n, k, ans, l, a[100010];
map<int, int>mp;
inline int pow(int y, int x){
	int s=1, t=y;
	while(x){if(x&1)s=(ll)s*t%mod; t=(ll)t*t%mod; x>>=1;}
	return s;
}
int main(){
	scanf("%d%d", &n, &k);
	l=0; mp.clear();
	for(int i=1; i*i<=n; i++)if(n%i==0){
		a[++l]=i; mp[i]=pow(k, (i+1)/2);
		if(i*i<n){a[++l]=n/i; mp[n/i]=pow(k, (n/i+1)/2);}
	}
	sort(a+1, a+1+l);
	for(int i=2; i<=l; i++){
		for(int j=1; j*j<=a[i]; j++)if(a[i]%j==0){
			mp[a[i]]=(mp[a[i]]-mp[j]+mod)%mod;
			if(j*j<a[i]&&j>=2)mp[a[i]]=(mp[a[i]]-mp[a[i]/j]+mod)%mod;
		}
	}
	ans=0;
	for(int i=1; i<=l; i++){
		if(a[i]&1)ans=(ans+(ll)a[i]*mp[a[i]]%mod)%mod;
		else ans=(ans+(ll)(a[i]/2)*mp[a[i]]%mod)%mod;
	}
	printf("%d", ans);
	return 0;
}