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 <stdio.h>
const int MOD = 1e9+7;
const int MAXV = 2048;
int ipow(int a, int b)
{
    int res = 1;
    while(b)
    {
        if(b&1) res = (1LL*res*a)%MOD;
        a = (1LL*a*a)%MOD;
        b /= 2;
    }
    return res;
}
int solve(int N, int K)
{
    int divisor[MAXV];
    int value[MAXV];
    int count = 0;
    {
        int i;
        for(i=1; i*i<N; ++i)
            if(N%i==0)
                divisor[count++] = i;
        if(i*i==N) divisor[count++] = i;        
        for(--i; i>0; --i)
            if(N%i==0)
                divisor[count++] = N/i;     
    }
    for(int i=0; i<count; ++i)
    {
        value[i] = ipow(K, (divisor[i]+1)/2);
        for(int j=0; j<i; ++j)
            if(divisor[i]%divisor[j] == 0)
            {
                value[i] -= value[j];
                if(value[i] < 0) value[i] += MOD;
            }
    }
    int ans = 0;
    for(int i=0; i<count; ++i)
    {
        if(divisor[i] % 2 == 0) divisor[i] /= 2;
        ans = (ans + 1LL * value[i] * divisor[i]) % MOD;
    }
    return ans;
}


int main()
{
    int N, K;
    scanf("%d%d", &N, &K);
    printf("%d\n", solve(N, K));

}