CF1316C.Primitive Primes

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.

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 $g(x)$ .

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.

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.