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 PyPy3 (7.3.0) (judge verdict: Accepted)

import sys
import math
from collections import defaultdict
from bisect import bisect_left, bisect_right

sys.setrecursionlimit(10**7)
def input():
    return sys.stdin.readline()[:-1]

mod = 10**9 + 7

def I(): return int(input())
def LI(): return list(map(int, input().split()))
def LIR(row,col):
    if row <= 0:
        return [[] for _ in range(col)]
    elif col == 1:
        return [I() for _ in range(row)]
    else:
        read_all = [LI() for _ in range(row)]
        return map(list, zip(*read_all))

#################

class OrderedSet:
    def __init__(self,n):
        self.c = {}
        self.r = {}
        #self.a_sort = sorted(list(set(a)))
        self.a_sort = list(range(n))
        for i,a in enumerate(self.a_sort):
            self.c[a] = i+1
            self.r[i+1] = a
        self.size = len(self.c)
        self.tree = [0]*(self.size + 1)
        self.num = [0]*self.size
        self.all_count = 0  # 全要素数

    # BITのsum関数
    def _sum(self,index):
        s = 0
        while index:
            s += self.tree[index]
            index -= index & (-index)
        return s

    # xをk個挿入する
    def insert(self,x,k=1):
        index = self.c[x]
        self.num[index-1] += k
        while index <= self.size:
            self.tree[index] += k
            index += index & (-index)
        self.all_count += k
    
    # xをk個削除する
    def erase(self,x,k=1):
        self.insert(x,k=-k)

    # 格納しているxの数を返す
    def count(self,x):
        return self.num[self.c[x]-1]

    # 下から数えてk番目の要素を返す（要素がk個未満のときNoneを返す）
    def get_kth_bottom(self,k):
        if k <= 0:
            return None
        elif self.all_count < k:
            return None
        else:
            i = 0
            s = 0
            step = 1 << (self.size.bit_length() - 1)
            while step:
                if i+step <= self.size and s+self.tree[i+step] < k:
                    i += step
                    s += self.tree[i]
                step >>= 1
            return self.r[i+1]
    
    # 上から数えてk番目の要素を返す（要素がk個未満のときNoneを返す）
    def get_kth_top(self,k):
        return self.get_kth_bottom(self.all_count-k+1)
    
    # x以上の最小要素を返す
    def lower_bound(self,x):
        if x in self.c.keys():
            p = self.c[x]-1
        else:
            p = bisect_left(self.a_sort,x)
        if p == 0:
            k = 1
        else:
            k = self._sum(p) + 1
        return self.get_kth_bottom(k)
    
    # x以下の最大要素を返す
    def upper_bound(self,x):
        if x in self.c.keys():
            p = self.c[x]
        else:
            p = bisect_right(self.a_sort,x)
        if p == 0:
            return None
        else:
            k = self._sum(p)
            return self.get_kth_bottom(k)

T = I()

ans = []
for _ in range(T):
    N = I()
    K,L,R = LIR(N,3)
    
    left = []    # 左側に入れたい要素
    right = []   # 右側に入れたい要素
    for i in range(N):
        large = max(L[i],R[i])
        small = min(L[i],R[i])
        gap = large-small
        if L[i] >= R[i]:
            left.append((gap,K[i]-1,large,small))
        else:
            right.append((gap,K[i],large,small))

    left.sort(key=lambda x:x[0], reverse=True)
    right.sort(key=lambda x:x[0], reverse=True)

    ordset = OrderedSet(N)
    ans_t = 0

    # まだ空いている場所をordsetに入れておく
    for i in range(N):
        ordset.insert(i)

    for i in range(len(left)):
        p = ordset.upper_bound(left[i][1])
        if p is None:
            p2 = ordset.upper_bound(len(left)-1)
            ordset.erase(p2)
            ans_t += left[i][3]
        elif p >= len(left):
            p2 = ordset.upper_bound(len(left)-1)
            ordset.erase(p2)
            ans_t += left[i][2]            
        else:
            ordset.erase(p)
            ans_t += left[i][2]

    for i in range(len(right)):
        p = ordset.lower_bound(right[i][1])
        if p is None:
            p2 = ordset.lower_bound(len(left))
            ordset.erase(p2)
            ans_t += right[i][3]
        else:
            ordset.erase(p)
            ans_t += right[i][2]
    
    ans.append(ans_t)

for ans_t in ans:
    print(ans_t)