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
// ====================================
//   author: M_sea
//   website: https://m-sea-blog.com/
// ====================================
#include <bits/stdc++.h>
#define file(x) freopen(#x".in","r",stdin); freopen(#x".out","w",stdout)
#define debug(...) fprintf(stderr,__VA_ARGS__)
using namespace std;
typedef long long ll;

int read() {
	int X=0,w=1; char c=getchar();
	while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
	while (c>='0'&&c<='9') X=X*10+c-'0',c=getchar();
	return X*w;
}

const int N=200000;
const int mod=1e9+7;
int qpow(int a,int b) { int c=1;
	for (;b;b>>=1,a=1ll*a*a%mod) if (b&1) c=1ll*c*a%mod;
	return c;
}

int n,k,fac[N],cnt=0,dp[N];

int main() {
	n=read(),k=read();
	for (int i=1;i*i<=n;++i) {
		if (n%i) continue;
		fac[++cnt]=i;
		if (i*i!=n) fac[++cnt]=n/i;
	}
	sort(fac+1,fac+cnt+1);
	for (int i=1;i<=cnt;++i) {
		dp[i]=qpow(k,(fac[i]+1)/2);
		for (int j=1;j<i;++j)
			if (fac[i]%fac[j]==0) dp[i]=(dp[i]-dp[j]+mod)%mod;
	}
	int ans=0;
	for (int i=1;i<=cnt;++i) {
		if (fac[i]&1) ans=(ans+1ll*dp[i]*fac[i])%mod;
		else ans=(ans+1ll*dp[i]*fac[i]/2)%mod;
	}
	printf("%d\n",ans);
	return 0;
}