Problem Statement

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 109+7?

Constraints

• 1≤N≤109
• 1≤K≤109

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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 109+7.

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Sample Input 1

Copy

4 2

Sample Output 1

Copy

6

The following six sequences can be obtained:

• (1,1,1,1)
• (1,1,2,2)
• (1,2,2,1)
• (2,2,1,1)
• (2,1,1,2)
• (2,2,2,2)

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Sample Input 2

Copy

1 10

Sample Output 2

Copy

10

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Sample Input 3

Copy

6 3

Sample Output 3

Copy

75

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Sample Input 4

Copy

1000000000 1000000000

Sample Output 4

Copy

875699961

用1-k填序列，序列可以左移n次变为回文

不考虑移位总共有k^(n/2)个回文，所以就是一个容斥问题了

若回文串的最小循环节长度为c,经过c次操作后,得到c个序列

当c为偶数时,重复数为一半,奇数时,无重复 

所以这个问题就解决了，其循环节只可能是其因子

#include 
       
        using namespace std;const int N=2010,MD=1e9+7;int v[N],f[N];int Pow(int x,int y){    int ans=1;    for(;y;x=x*1LL*x%MD,y>>=1)if(y&1)ans=1LL*ans*x%MD;    return ans;}int main(){    int n,k,tot=0,ans=0,i;    cin>>n>>k;    for(i=1; i*i