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<cctype>
#include<cstring>
#include<cmath>
#include<algorithm>
#define max(a,b) ((a)>(b)?(a):(b))
#define reg register
typedef long long ll;
int n,k,cnt,fac[2005];
ll f[2005],ans;
const int mod=1e9+7;
inline ll qpow(ll a,ll b){
	reg ll res=1;
	while(b){
		if(b&1)res=res*a%mod;
		a=a*a%mod;b>>=1;
	}
	return res;
}
int main(){
	scanf("%d%d",&n,&k);
	for(reg int i=1;i<=sqrt(n);i++)
		if(!(n%i)){
			fac[++cnt]=i;
			if(i!=(n/i))fac[++cnt]=n/i;
		}
	std::sort(fac+1,fac+1+cnt);
	for(reg int i=1;i<=cnt;i++){
		f[i]=qpow(k,(fac[i]+1)/2);
		for(reg int j=1;j<i;j++)
			if(!(fac[i]%fac[j]))
				f[i]=(f[i]-f[j]+mod)%mod;
		if(fac[i]&1)ans=(ans+f[i]*fac[i]%mod)%mod;
		else ans=(ans+(f[i]*fac[i]>>1)%mod)%mod;
	}
	printf("%lld\n",ans);
    return 0;
}