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 ll long long
#define mod 1000000007
using namespace std;

int n,m,cnt,sum,a[100005],ans[100005]; bool flag;
int ksm(int x,int y){
	int z=1; for (; y; y>>=1,x=(ll)x*x%mod) if (y&1) z=(ll)z*x%mod;
	return z;
}
int main(){
	scanf("%d%d",&n,&m);
	int i,j; 
	if (n&1) flag=1; else n/=2;
	for (i=1; i*i<=n; i++) if (!(n%i)){
		a[++cnt]=i;
		if (i*i<n) a[++cnt]=n/i;
	}
	sort(a+1,a+cnt+1);
	for (i=1; i<=cnt; i++){
		ans[i]=(ans[i]+ksm(m,(flag)?(a[i]+1>>1):a[i]))%mod;
		sum=(sum+(ll)ans[i]*a[i])%mod;
		for (j=i+1; j<=cnt; j++) if (!(a[j]%a[i])) ans[j]=(ans[j]+mod-ans[i])%mod;
	}
	printf("%d\n",sum);	
	return 0;
}
