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 <cstring>
#include <string>
#include <iostream>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
const ll MOD=1e9+7;
ll powmod(ll a, ll k){
    ll ap=a, ans=1;
    while(k>0){
        if(k%2==1){
            ans*=ap;
            ans%=MOD;
        }
        ap=ap*ap;
        ap%=MOD;
        k/=2;
    }
    return ans;
}
int main()
{
	ll n, k;
  cin>>n>>k;
  vector<ll> div;
  for(ll i=1; i*i<=n; i++){
    if(n%i==0){
      div.push_back(i);
      if(i*i<n) div.push_back(n/i);
    }
  }
  sort(div.begin(), div.end());
  ll ct1[1350], ct[1350];
  for(int i=0; i<div.size(); i++){
    ct1[i]=powmod(k, (div[i]+1)/2);
    ct[i]=ct1[i];
    for(int j=0; j<i; j++){
      if(div[i]%div[j]==0){
        ct[i]-=ct[j];
        ct[i]+=MOD;
        ct[i]%=MOD;
      }
    }
  }
  ll ans=0;
  for(int i=0; i<div.size(); i++){
    if(div[i]%2!=0) ans+=(ct[i]*div[i]%MOD);
    else ans+=(ct[i]*div[i]%MOD*(ll)(5e8+4)%MOD);
    ans%=MOD;
  }
  cout<<ans<<endl;
	return 0;
}