# The Destruction of the Universe (Easy Version)

## Problem Description

This is the easy version of the problem. In this version, $ n \leq 5000 $ . You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $ 1 $ unit of time to destroy every vulnerable planet in the universe!

There are $ n $ planets in the universe. Each planet has an interval of vulnerability $ [l, r] $ , during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $ 1 $ unit.

Specifically, suppose the expansion is performed on planet $ p $ with interval of vulnerability $ [l_p, r_p] $ . Then, the resulting interval of vulnerability may be either $ [l_p - 1, r_p] $ or $ [l_p, r_p + 1] $ .

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $ 998\,244\,353 $ .

## Input Format

The first line contains an integer $ t $ ( $ 1 \leq t \leq 1000 $ ) — the number of test cases.

The first line of each test case contains an integer $ n $ ( $ 1 \leq n \leq 5000 $ ) — the number of planets in the universe.

The following $ n $ lines contain two integers $ l_i $ and $ r_i $ ( $ 1 \leq l_i \leq r_i \leq n $ ) — the initial interval of vulnerability of the $ i $ -th planet.

It is guaranteed that the sum of $ n $ does not exceed $ 5000 $ over all test cases.

## Output Format

For each test case, output an integer — the sum of scores to destroy all non-empty subsets of the planets in the universe, modulo $ 998\,244\,353 $ .

## Sample #1

### Sample Input #1

```
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
```

### Sample Output #1

```
5
6
24
```

## Hint

In the first testcase, there are seven non-empty subsets of planets we must consider:

- For each of the subsets $ \{[1,1]\}, \{[2,3]\}, \{[3,3]\} $ , the score is $ 0 $ .
- For the subset $ \{[2,3], [3,3]\} $ , the score is $ 0 $ , because the point $ 3 $ is already contained in both planets' interval of vulnerability.
- For the subset $ \{[1,1], [2,3]\} $ , the score is $ 1 $ . By using one operation on changing the interval of vulnerability of the second planet to be $ [1,3] $ , the two planets now both have the point $ 1 $ in their interval.
- For the subset $ \{[1,1], [3,3]\} $ , the score is $ 2 $ . By using two operations on changing the interval of vulnerability of the first planet to be $ [1,3] $ , the two planets now both have the point $ 3 $ in their interval.
- For the subset $ \{[1,1], [2,3], [3,3]\} $ , the score is $ 2 $ . By using one operation on changing the interval of vulnerability of the first planet to be $ [1,2] $ and one operation on changing the interval of vulnerability of the third planet to $ [2,3] $ , all three planets will have the point $ 2 $ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $ 0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5 $ .

## Limit



Time Limit
4.00s

Memory Limit
500.00MB

Tags: combinatorics, greedy, math
Difficulty: 2900