D. Ceil Divisions
brute force; constructive algorithms; math; number theory
You have an array \(a_1, a_2, \dots, a_n\) where \(a_i = i\).In one step, you can choose two indices \(x\) and \(y\) (\(x \neq y\)) and set \(a_x = \left\lceil \frac{a_x}{a_y} \right\rceil\) (ceiling function).Your goal is to make array \(a\) consist of \(n - 1\) ones and \(1\) two in no more than \(n + 5\) steps. Note that you don't have to minimize the number of steps.
The first line contains a single integer \(t\) (\(1 \le t \le 1000\)) — the number of test cases.The first and only line of each test case contains the single integer \(n\) (\(3 \le n \le 2 \cdot 10^5\)) — the length of array \(a\).It's guaranteed that the sum of \(n\) over test cases doesn't exceed \(2 \cdot 10^5\).
For each test case, print the sequence of operations that will make \(a\) as \(n - 1\) ones and \(1\) two in the following format: firstly, print one integer \(m\) (\(m \le n + 5\)) — the number of operations; next print \(m\) pairs of integers \(x\) and \(y\) (\(1 \le x, y \le n\); \(x \neq y\)) (\(x\) may be greater or less than \(y\)) — the indices of the corresponding operation.It can be proven that for the given constraints it's always possible to find a correct sequence of operations.
In the first test case, you have array \(a = [1, 2, 3]\). For example, you can do the following: choose \(3\), \(2\): \(a_3 = \left\lceil \frac{a_3}{a_2} \right\rceil = 2\) and array \(a = [1, 2, 2]\); choose \(3\), \(2\): \(a_3 = \left\lceil \frac{2}{2} \right\rceil = 1\) and array \(a = [1, 2, 1]\). You've got array with \(2\) ones and \(1\) two in \(2\) steps.In the second test case, \(a = [1, 2, 3, 4]\). For example, you can do the following: choose \(3\), \(4\): \(a_3 = \left\lceil \frac{3}{4} \right\rceil = 1\) and array \(a = [1, 2, 1, 4]\); choose \(4\), \(2\): \(a_4 = \left\lceil \frac{4}{2} \right\rceil = 2\) and array \(a = [1, 2, 1, 2]\); choose \(4\), \(2\): \(a_4 = \left\lceil \frac{2}{2} \right\rceil = 1\) and array \(a = [1, 2, 1, 1]\).