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
/*
数据不清空，爆零两行泪。
多测不读完，爆零两行泪。
边界不特判，爆零两行泪。
贪心不证明，爆零两行泪。
D P 顺序错，爆零两行泪。
大小少等号，爆零两行泪。
变量不统一，爆零两行泪。
越界不判断，爆零两行泪。
调试不注释，爆零两行泪。
溢出不 l l，爆零两行泪。
*/
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
const int mod=1000000007;
int qpow(int x,int y){
	int res=1;
	while(y){
		if(y&1)(res*=x)%=mod;
		(x*=x)%=mod;
		y>>=1;
	}
	return res;
}
int n,m;
vector<int> d;
vector<int> dp;
signed main(){
	cin>>n>>m;
	for(int i=1;i*i<=n;i++)if(n%i==0)
		d.pb(i),d.pb(n/i);
	sort(d.begin(),d.end());
	d.resize(unique(d.begin(),d.end())-d.begin());
	int ans=0;
	for(int i=0;i<d.size();i++){
		dp.pb(qpow(m,(d[i]+1)/2));
		for(int j=0;j<i;j++)if(d[i]%d[j]==0)(((dp[i]-=dp[j])%=mod)+=mod)%=mod;
		(ans+=dp[i]*(d[i]&1?d[i]:d[i]/2))%=mod;
//		printf("i=%lld ans=%lld\n",i,ans);
	}
	cout<<ans;
	return 0;
}
/*1
4 2
*/
/*2
1 10
*/
/*3
6 3
*/
/*4
1000000000 1000000000
*/