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>
#define mod 1000000007
#define int long long
using namespace std;
int dp[2010],cnt,n,k,a[2010],ans;
int fastpow(int x,int y)
{
    int sum=1;
    while(y)
    {
        if(y&1)
        {
            sum=(1ll*sum*x)%mod;
        }
        x=(1ll*x*x)%mod;
        y>>=1;
    }
    return sum;
}
signed main()
{
    scanf("%lld%lld",&n,&k);
    for(int i=1;i*i<=n;i++)//枚举因子
    {
        if(n%i==0)
        {
            a[++cnt]=i;
            if(i*i!=n)
            {
                a[++cnt]=n/i;
            }
        }
    }
    sort(a+1,a+cnt+1);
    for(int i=1;i<=cnt;i++)
    {
        dp[i]=fastpow(k,(a[i]+1)/2);//有多少个回文串
        for(int j=1;j<i;j++)
        {
            if(a[i]%a[j]==0)//去重
            {
                dp[i]=(dp[i]-dp[j]+mod)%mod;
            }
        }
        if(a[i]&1)//统计答案
        {
            ans=(ans+dp[i]*a[i])%mod;
        }else{
            ans=(ans+dp[i]*(a[i]/2))%mod;
        }
    }
    printf("%lld\n",ans);
    return 0;
}