It is Professor R's last class of his teaching career. Every time Professor R taught a class, he gave a special problem for the students to solve. You being his favourite student, put your heart into solving it one last time.

You are given two polynomials $f(x) = a_0 + a_1x + \dots + a_{n-1}x^{n-1}$ and $g(x) = b_0 + b_1x + \dots + b_{m-1}x^{m-1}$, with positive integral coefficients. It is guaranteed that the cumulative GCD of the coefficients is equal to $1$ for both the given polynomials. In other words, $gcd(a_0, a_1, \dots, a_{n-1}) = gcd(b_0, b_1, \dots, b_{m-1}) = 1$. Let $h(x) = f(x)\cdot g(x)$. Suppose that $h(x) = c_0 + c_1x + \dots + c_{n+m-2}x^{n+m-2}$. 

You are also given a prime number $p$. Professor R challenges you to find any $t$ such that $c_t$ isn't divisible by $p$. He guarantees you that under these conditions such $t$ always exists. If there are several such $t$, output any of them.

As the input is quite large, please use fast input reading methods.


-----Input-----

The first line of the input contains three integers, $n$, $m$ and $p$ ($1 \leq n, m \leq 10^6, 2 \leq p \leq 10^9$),  — $n$ and $m$ are the number of terms in $f(x)$ and $g(x)$ respectively (one more than the degrees of the respective polynomials) and $p$ is the given prime number.

It is guaranteed that $p$ is prime.

The second line contains $n$ integers $a_0, a_1, \dots, a_{n-1}$ ($1 \leq a_{i} \leq 10^{9}$) — $a_i$ is the coefficient of $x^{i}$ in $f(x)$.

The third line contains $m$ integers $b_0, b_1, \dots, b_{m-1}$ ($1 \leq b_{i} \leq 10^{9}$)  — $b_i$ is the coefficient of $x^{i}$ in $g(x)$.


-----Output-----

Print a single integer $t$ ($0\le t \le n+m-2$)  — the appropriate power of $x$ in $h(x)$ whose coefficient isn't divisible by the given prime $p$. If there are multiple powers of $x$ that satisfy the condition, print any.


-----Examples-----
Input
3 2 2
1 1 2
2 1

Output
1

Input
2 2 999999937
2 1
3 1

Output
2


-----Note-----

In the first test case, $f(x)$ is $2x^2 + x + 1$ and $g(x)$ is $x + 2$, their product $h(x)$ being $2x^3 + 5x^2 + 3x + 2$, so the answer can be 1 or 2 as both 3 and 5 aren't divisible by 2.

In the second test case, $f(x)$ is $x + 2$ and $g(x)$ is $x + 3$, their product $h(x)$ being $x^2 + 5x + 6$, so the answer can be any of the powers as no coefficient is divisible by the given prime.

Difficulty: interview

Starter code:

Solutions:
from random import randint
import sys

n, m, p = list(map(int, input().strip().split()))

f = list(map(int, input().strip().split()))
g = list(map(int, input().strip().split()))

for i in range(len(f)):
    if f[i] % p != 0:
        break

j = 0
while g[j] % p == 0:
    j += 1

print(i+j)






import sys
input = lambda: sys.stdin.readline().rstrip()
N, M, P = list(map(int, input().split()))
A = [int(a) for a in input().split()]
B = [int(a) for a in input().split()]
for i, a in enumerate(A):
    if a % P:
        for j, b in enumerate(B):
            if b % P:
                print(i+j)
                return