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>
using namespace std;

using ll=long long;
const ll MOD=1e9+7;

ll powm(ll a,ll b){
    ll res=1;
    while(b){
        if(b&1) res*=a;
        a*=a;
        a%=MOD;
        res%=MOD;
        b>>=1;
    }
    return res;
}

ll mod_inv(ll x){
    return powm(x,MOD-2);
}

int main(){
    ll n,k;
    cin>>n>>k;
    set<int> divisor;
    for(int i=1;i<1e5;i++){
        if(n%i==0){
            divisor.insert(i);
            divisor.insert(n/i);
        }
    }
    vector<ll> div(divisor.begin(),divisor.end());

    vector<ll> dp(div.size());
    ll res=0;
    
    for(int i=0;i<div.size();i++){
        ll cnt=powm(k,div[i]/2+(ll)(div[i]%2!=0));
        cnt=(cnt+MOD)%MOD;
        for(int j=0;j<i;j++){
            if(div[i]%div[j]==0) cnt-=dp[j];
            cnt=(cnt%MOD+MOD)%MOD;
        }
        dp[i]=cnt;
        if(div[i]%2==0){
            res+=(((cnt*mod_inv(2))%MOD)*div[i])%MOD;
        }
        else{
            res+=(cnt*div[i])%MOD;
        }
        res%=MOD;
    }
    cout<<res<<endl;

    return 0;
}
