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 pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const ll MOD = 1000000007;

ll mod_pow(ll x, ll k) {
	ll res = 1;
	for (; k; k /= 2, x = x * x % MOD) {
		if (k & 1) res = res * x % MOD;
	}
	return res;
}

int N, K;
ll ret;

int main() {
	cin >> N >> K;

	vi dv;

	for (int i = 1; i * i <= N; ++i) {
		if (N % i == 0) {
			dv.pb(i);
			if (i * i != N) {
				dv.pb(N / i);
			}
		}
	}
	sort(ALL(dv));

	vector<ll> vec(dv.size());

	ll ret = 0;
	rep(i, dv.size()) {
		ll t = mod_pow(K, (dv[i] + 1) / 2);
		rep(j, i) {
			if (dv[i] % dv[j] == 0) {
				t = (t - vec[j]);
				if (t < 0) t += MOD;
			}
		}
		vec[i] = t;

		if (dv[i] & 1) {
			ret = (ret + vec[i] * dv[i]) % MOD;
		} else {
			ret = (ret + vec[i] * dv[i] / 2) % MOD;
		}
	}
	cout << ret << endl;

	return 0;
}