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
import sys
def MI(): return map(int,sys.stdin.readline().rstrip().split())

N,K = MI()
mod = 10**9+7

def divisor(n):  # nの約数のリスト
    res = []
    for i in range(1,int(n**.5)+1):
        if n % i == 0:
            res.append(i)
            if i != n//i:
                res.append(n//i)
    return res

A = divisor(N)

d = {}  # d[i] = iの約数のリスト(iはNの約数)
for a in A:
    d[a] = divisor(a)

prime = []  # Nの素因数のリスト
for i in range(2,int(N**.5)+1):
    if N % i == 0:
        prime.append(i)
        while N % i == 0:
            N //= i
if N != 1:
    prime.append(N)

mu = {}  # mu[i] = μ(i) (iはNの約数)
for a in A:
    b = a
    r = 1
    for p in prime:
        if b % p == 0:
            r *= -1
            if b//p % p == 0:
                r = 0
                break
    mu[a] = r


ans = 0
for a in A:
    for b in d[a]:
        if a % 2 == 0:
            ans += mu[a//b]*pow(K,(b+1)//2,mod)*(a//2)
            ans %= mod
        else:
            ans += mu[a//b]*pow(K,(b+1)//2,mod)*a
            ans %= mod

print(ans)