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;

typedef long long ll;
const int MOD = 1000000007;

int n, k, ans;
vector<int> fac, dp;

inline int pwr(int a, int b) {
    int c = 1;
    while (b) {
        if (b & 1) c = (ll)c * a % MOD;
        b >>= 1; a = (ll)a * a % MOD;
    }
    return c;
}
inline void add(int &x, int a) {
    x = (x + a >= MOD) ? x + a - MOD : x + a;
}
inline void sub(int &x, int a) {
    x = (x < a) ? x - a + MOD : x - a;
}

int main(void) {
    cin >> n >> k;
    int m = sqrt(n);
    for (int i = 1; i <= m; i++)
        if (n % i == 0) {
            fac.push_back(i);
            fac.push_back(n / i);
        }
    sort(fac.begin(), fac.end());
    fac.erase(unique(fac.begin(), fac.end()), fac.end());
    dp.resize(fac.size());
    for (int i = 0; i < fac.size(); i++) {
        dp[i] = pwr(k, (fac[i] + 1) / 2);
        for (int j = 0; j < i; j++)
            if (fac[i] % fac[j] == 0)
                sub(dp[i], dp[j]);
        if (fac[i] & 1) add(ans, (ll)dp[i] * fac[i] % MOD);
        else add(ans, (ll)dp[i] * fac[i] / 2 % MOD);
    }
    cout << ans << endl;
    return 0;
}
