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;

const int Nmax = 3005, Mod = 1e9 + 7;
typedef long long ll;

void add_to(int &x, int y) { x+=y; if(x>=Mod) x-=Mod; }
int inm(int x, int y) { return (ll)x*y%Mod; }

int N, K, i, j, ans = 0, dp[Nmax];
vector<int> v, v2;

int power(int x, int y)
{
    int ans = 1;
    while(y)
    {
        if(y&1) ans = inm(ans, x);
        x = inm(x, x); y >>= 1;
    }
    return ans;
}

int main()
{
///    freopen("input", "r", stdin);
    cin.sync_with_stdio(false);

    cin >> N >> K;
    for(i=1; i*i<=N; ++i)
        if(N % i == 0)
        {
            v.push_back(i);
            if(i*i < N) v2.push_back(N/i);
        }

    reverse(v2.begin(), v2.end());
    for(auto it : v2) v.push_back(it);

    for(i=0; i<v.size(); ++i)
    {
        dp[i] = power( K, (v[i]+1)/2 );

        for(j=0; j<i; ++j)
            if(v[i] % v[j] == 0)
                add_to(dp[i], Mod - dp[j]);

        add_to(ans, inm( dp[i], v[i] / (1<<(v[i]%2==0)) ) );
    }

    cout << ans << '\n';
    return 0;
}
