## Problem

Chef Ada is preparing NN dishes (numbered 11 through NN). For each valid ii, it takes CiCi minutes to prepare the ii-th dish. The dishes can be prepared in any order.

Ada has a kitchen with two identical burners. For each valid ii, to prepare the ii-th dish, she puts it on one of the burners and after CiCi minutes, removes it from this burner; the dish may not be removed from the burner before those CiCi minutes pass, because otherwise it cools down and gets spoiled. Any two dishes may be prepared simultaneously, however, no two dishes may be on the same burner at the same time. Ada may remove a dish from a burner and put another dish on the same burner at the same time.

What is the minimum time needed to prepare all dishes, i.e. reach the state where all dishes are prepared?

### Input

- The first line of the input contains a single integer TT denoting the number of test cases. The description of TT test cases follows.
- The first line of each test case contains a single integer NN.
- The second line contains NN space-separated integers C1,C2,…,CNC1,C2,…,CN.

### Output

For each test case, print a single line containing one integer ― the minimum number of minutes needed to prepare all dishes.

### Constraints

- 1≤T≤1,0001≤T≤1,000
- 1≤N≤41≤N≤4
- 1≤Ci≤51≤Ci≤5 for each valid ii

### Subtasks

**Subtask #1 (1 points):** C1=C2=…=CNC1=C2=…=CN

**Subtask #2 (99 points):** original constraints

### Sample Input 1

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

### Sample Output 1

```
4
3
7
```

### Explanation

**Example case 1:** Place the first two dishes on the burners, wait for two minutes, remove both dishes and prepare the last one on one burner.

**Example case 2:** Place the first and third dish on the burners. When the first dish is prepared, remove it and put the second dish on the same burner.

**Example case 3:** Place the third and fourth dish on the burners. When the third dish is prepared, remove it and put the second dish on the same burner. Similarly, replace the fourth dish (when it is prepared) by the first dish on the other burner

## Ada and Dishes – CodeChef Solution in JAVA

import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int T = sc.nextInt(); for (int tc = 0; tc < T; ++tc) { int N = sc.nextInt(); int[] C = new int[N]; for (int i = 0; i < C.length; ++i) { C[i] = sc.nextInt(); } System.out.println(solve(C)); } sc.close(); } static int solve(int[] C) { int result = Integer.MAX_VALUE; for (int code = 0; code < 1 << C.length; ++code) { int[] times = new int[2]; for (int i = 0; i < C.length; ++i) { times[((code & (1 << i)) == 0) ? 0 : 1] += C[i]; } result = Math.min(result, Math.max(times[0], times[1])); } return result; } }

** Disclaimer: The above Problem ( Ada and Dishes) is generated by** CodeChef but the solution is provided by

**Chase2learn**. This tutorial is only for

**Educational**and

**Learning**purpose.