Problem:
Score : 500 points

Problem Statement

We have N camels numbered 1,2,\ldots,N.
Snuke has decided to make them line up in a row.

The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.

Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.

Solve this problem for each of the T test cases given.

Constraints

All values in input are integers.

1 \leq T \leq 10^5

1 \leq N \leq 2 \times 10^{5}

1 \leq K_i \leq N

1 \leq L_i, R_i \leq 10^9

The sum of values of N in each input file is at most 2 \times 10^5.

Input

Input is given from Standard Input in the following format:

T
\mathrm{case}_1
\vdots
\mathrm{case}_T

Each case is given in the following format:

N
K_1 L_1 R_1
\vdots
K_N L_N R_N

Output

Print T lines. The i-th line should contain the answer to the i-th test case.

Sample Input 1

3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34

Sample Output 1

25
221
1354

In the first test case, it is optimal to line up the camels in the order 2, 1.

Camel 1 is not the frontmost camel, so its happiness will be 10.

Camel 2 is among the two frontmost camels, so its happiness will be 15.

In the second test case, it is optimal to line up the camels in the order 2, 1, 3.

Camel 1 is among the two frontmost camels, so its happiness will be 93.

Camel 2 is the frontmost camel, so its happiness will be 71.

Camel 3 is among the three frontmost camels, so its happiness will be 57.

Solution in OCaml (4.10.0) (judge verdict: Accepted)

Scanf.scanf "%d" (fun t ->
    let module S = Set.Make (struct type t = int * int let compare = compare end) in
    for i = 1 to t do
        Scanf.scanf " %d" (fun n ->
            let klr = Array.init n (fun _ -> Scanf.scanf " %d %d %d" (fun k l r -> k, l, r)) in
            let sum = Array.fold_left (fun acc (_, l, r) -> acc + min l r) 0 klr in

            let rec loop i left right =
                if i = n then Array.of_list left, Array.of_list right else
                    let k, l, r = klr.(i) in
                    if l > r then loop (i + 1) ((k, l - r) :: left) right
                             else loop (i + 1) left ((n - k, r - l) :: right)
            in
            let left, right = loop 0 [] [] in
            Array.sort compare left;
            Array.sort compare right;
            let proc a =
                let len = Array.length a in
                let rec loop i j acc set ct =
                    if i = len then acc else
                        let k, dif =  a.(i) in
                        if k > j then loop i (j + 1) acc set ct else
                            let set = S.add (dif, i) set in
                            let ct = ct + 1 in
                            let acc = acc + dif in
                            if ct <= k then loop (i + 1) j acc set ct else
                                let ((dif, _) as m) = S.min_elt set in
                                let set = S.remove m set in
                                let ct = ct - 1 in
                                let acc = acc - dif in
                                loop (i + 1) j acc set ct
                in
                loop 0 0 0 S.empty 0
            in
            Printf.printf "%d\n" @@ sum + proc left + proc right
        )
    done
)