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 <algorithm>
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
#define int long long
#define mod (int)1000000007
#define N 64000
vector<int> vec;
int tot[N];
int pow(int base,int up)
{
    int ans=1;
    while(up)
    {
        if(up&1) ans*=base,ans%=mod;
        base*=base;
        base%=mod;
        up>>=1;
    }
    return ans;
}
signed main()
{
    int n,k,len;
    cin>>n>>k;
    len=sqrt(n);
    for(int i=1;i<=len;i++) if(n%i==0)//获取所有可能的循环节长度
    {
        vec.push_back(i);
        if(i!=n/i) vec.push_back(n/i);
    }
    sort(vec.begin(), vec.end());
    int ans=0;
    for(int i=0;i<vec.size();i++)
    {
        tot[i]=pow(k,(vec[i]+1)/2);//循环节长度为vec[i]时循环节的数量。因为是回文串，所以只需要知道前面半截即可确定出整个回文串，下面也是同理
        for(int j=0;j<i;j++) if(vec[i]%vec[j]==0)
        tot[i]-=tot[j],tot[i]=(tot[i]+mod)%mod;//去除【长度为vec[i]的循环节】内部含有循环节的情况
        if(vec[i]&1) ans+=tot[i]*vec[i],ans%=mod;
        else ans+=tot[i]*(vec[i]/2),ans%=mod;
    }
    cout<<ans;
}