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;

#define FOR(i, x, y) for(int i = (x); i < (y); ++i)
#define REP(i, x, y) for(int i = (x); i <= (y); ++i)
#define MP make_pair
#define PB push_back
#define PH push
#define fst first
#define snd second
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<int, int> pii;

const int magic = 1e5 + 5;
const int INF = 1e9 + 7;

int n, k;
int f[magic];
int ans;
vector<int> vec;

inline int qpow(ll x, int y){
	ll ret = 1;
	for(; y; y >>= 1){
		if(y & 1) (ret *= x) %= INF;
		(x *= x) %= INF;
	}
	return ret;
}

int main(){
	scanf("%d%d", &n, &k);
	for(int i = 1; i * i <= n; ++i) if(!(n % i)){
		vec.PB(i);
		if(n / i != i) vec.PB(n / i);
	}
	sort(vec.begin(), vec.end());
	FOR(i, 0, vec.size()){
		FOR(j, 0, i) if(!(vec[i] % vec[j])) ((f[i] -= f[j]) += INF) %= INF;
		(f[i] += qpow(k, (vec[i] + 1) / 2)) %= INF;
		if(vec[i] & 1) (ans += (1ll * vec[i] * f[i]) % INF) %= INF;
		else (ans += (1ll * vec[i] / 2 * f[i]) % INF) %= INF;
	}
	printf("%d\n", ans);
	return 0;
}