CF1740A.Factorise N+M

Pak Chanek has a prime number$^\dagger$ $n$. Find a prime number $m$ such that $n + m$ is not prime.

$^\dagger$ A prime number is a number with exactly $2$ factors. The first few prime numbers are $2,3,5,7,11,13,\ldots$. In particular, $1$ is not a prime number.

帕克·查内克有一个质数 $n$。请找出一个质数 $m$，使得 $n + m$ 不是质数。

$^\dagger$ 质数是指恰好有 $2$ 个正因数的正整数。前几个质数是 $2,3,5,7,11,13,\ldots$。特别地，$1$ 不是质数。

Each test contains multiple test cases. The first line contains an integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases. The following lines contain the description of each test case.

The only line of each test case contains a prime number $n$ ($2 \leq n \leq 10^5$).

每个测试包含多个测试用例。第一行包含一个整数 $t$（$1 \leq t \leq 10^4$），表示测试用例的数量。接下来的各行描述每个测试用例。

每个测试用例仅有一行，包含一个质数 $n$（$2 \leq n \leq 10^5$）。

For each test case, output a line containing a prime number $m$ ($2 \leq m \leq 10^5$) such that $n + m$ is not prime. It can be proven that under the constraints of the problem, such $m$ always exists.

If there are multiple solutions, you can output any of them.

对于每个测试用例，输出一行，包含一个素数 $m$（$2 \leq m \leq 10^5$），使得 $n + m$ 不是素数。在本题的约束条件下，可以证明这样的 $m$ 总是存在的。

如果存在多个解，你可以输出其中任意一个。

In the first test case, $m = 2$, which is prime, and $n + m = 7 + 2 = 9$, which is not prime.

In the second test case, $m = 7$, which is prime, and $n + m = 2 + 7 = 9$, which is not prime.

In the third test case, $m = 47837$, which is prime, and $n + m = 75619 + 47837 = 123456$, which is not prime.

在第一个测试用例中，$m = 2$ 是质数，而 $n + m = 7 + 2 = 9$ 不是质数。

在第二个测试用例中，$m = 7$ 是质数，而 $n + m = 2 + 7 = 9$ 不是质数。

在第三个测试用例中，$m = 47837$ 是质数，而 $n + m = 75619 + 47837 = 123456$ 不是质数。