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 syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<P,int> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<55;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

ll Pow(ll n,ll p){
	ll r=1;
	for(;p>0;p>>=1){
		if(p&1) r=(r*n)%mod;
		n=(n*n)%mod;
	}
	return r;
}

int n,k;
vi a;

int main(){
	cin>>n>>k;
	for(int i=1;i*i<=n;i++) if(n%i==0){
		a.push_back(i);
		if(n/i!=i) a.push_back(n/i);
	}
	sort(a.begin(),a.end());
	int N=a.size();
	vl b(N);
	ll res=0;
	for(int i=0;i<N;i++){
		ll d=a[i],tmp=Pow(k,(d+1)/2);
		for(int j=0;j<i;j++) if(d%a[j]==0) tmp=(tmp-b[j]+mod)%mod;
		b[i]=tmp;
		(res+=b[i]*(d%2==0?d/2:d))%=mod;
	}
	cout<<res<<endl;
}