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>
#include<cmath>
#define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
#define mod 1000000007
#define maxn 10010

using namespace std;
typedef long long llg;

int n,k;
int a[maxn],la;
llg f[maxn],ans;

llg mi(llg a,int b){
	llg s=1;
	while(b){
		if(b&1) s=s*a%mod;
		a=a*a%mod; b>>=1;
	}
	return s;
}

int main(){
	scanf("%d %d",&n,&k);
	for(int i=1;i*i<=n;i++)
		if(n%i==0) a[++la]=i,a[++la]=n/i;
	sort(a+1,a+la+1); la=unique(a+1,a+la+1)-a-1;
	for(int i=1;i<=la;i++){
		f[i]+=mi(k,(a[i]+1)/2); f[i]%=mod;
		if(f[i]<0) f[i]+=mod;
		for(int j=i+1;j<=la;j++)
			if(a[j]%a[i]==0) f[j]-=f[i];
		if(~a[i]&1) a[i]>>=1;
		ans+=f[i]*a[i]; ans%=mod;
	}
	printf("%lld",ans);
	return 0;
}