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;

LL N,K;
const LL MOD = 1000000007ll;

LL m_pow(LL a,LL x)
{
    if(x == 0)return 1;
    LL half = m_pow(a,x/2);
    half *= half;
    half %= MOD;
    return (half * (x % 2 ? a : 1))%MOD;
}

map<LL,LL>memo;
LL dp(LL d)
{
    if(memo.count(d))return memo[d];
    LL res = m_pow(K,(d+1)/2);
    vector<int>div;
    for(int i = 1;i * i <= d;++i)
    {
        if(d % i == 0)
        {
            div.push_back(i);
            if(d/i != i)div.push_back(d/i);
        }
    }
    for(int dash : div)
    {
        if(dash == d)continue;
        res -= dp(dash);
        res += MOD;
        res %= MOD;
    }
    return memo[d] = res;
}

int main()
{
    cin >> N >> K;
    dp(N);
    LL ans = 0;
    for(auto p : memo)
    {
        LL index = p.first;
        LL value = p.second;
        if(index % 2 == 0)
        {
            value *= index / 2;
        }
        else
        {
            value *= index;
        }
        /*
        cerr << index << endl;
        cerr << value << endl;
        //*/
        value %= MOD;
        ans += value;
        ans %= MOD;
    }
    cout << ans << endl;
    return 0;
}