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

Sample Input 1

4 2

Sample Output 1

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)

Sample Input 2

1 10

Sample Output 2

10

Sample Input 3

6 3

Sample Output 3

75

Sample Input 4

1000000000 1000000000

Sample Output 4

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 <bits/stdc++.h>

using namespace std;

const int N=,MD=1e9+;

int v[N],f[N];

int Pow(int x,int y)

{

    int ans=;

    for(;y;x=x*1LL*x%MD,y>>=)if(y&)ans=1LL*ans*x%MD;

    return ans;

}

int main()

{

    int n,k,tot=,ans=,i;

    cin>>n>>k;

    for(i=; i*i<n; i++)

        if(n%i==)v[tot++]=i,v[tot++]=n/i;

    if(i*i==n)v[tot++]=i;

    sort(v,v+tot);

    for(int i=; i<tot; i++)

    {

        f[i]=Pow(k,(v[i]+)/);

        for(int j=; j<i; j++)if(v[i]%v[j]==) f[i]=(f[i]-f[j])%MD;

        if(v[i]&) ans=(ans+1LL*v[i]*f[i])%MD;

        else ans=(ans+v[i]*1LL*Pow(,MD-)%MD*f[i])%MD;

    }

    cout<<(ans+MD)%MD;

}