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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<ll, ll>;
using Pld = pair<ld, ld>;
using Vec = vector<ll>;
using VecP = vector<P>;
using VecB = vector<bool>;
using VecC = vector<char>;
using VecD = vector<ld>;
using VecS = vector<string>;
using Graph = vector<VecP>;
template <typename T>
using Vec1 = vector<T>;
template <typename T>
using Vec2 = vector<Vec1<T> >;
#define REP(i, m, n) for(ll i = (m); i < (n); ++i)
#define REPN(i, m, n) for(ll i = (m); i <= (n); ++i)
#define REPR(i, m, n) for(ll i = (m)-1; i >= (n); --i)
#define REPNR(i, m, n) for(ll i = (m); i >= (n); --i)
#define rep(i, n) REP(i, 0, n)
#define repn(i, n) REPN(i, 1, n)
#define repr(i, n) REPR(i, n, 0)
#define repnr(i, n) REPNR(i, n, 1)
#define all(s) (s).begin(), (s).end()
#define pb push_back
#define fs first
#define sc second
template <typename T>
bool chmax(T &a, const T b){if(a < b){a = b; return true;} return false;}
template <typename T>
bool chmin(T &a, const T b){if(a > b){a = b; return true;} return false;}
template <typename T>
ll pow2(const T n){return (1LL << n);}
template <typename T>
void cosp(const T n){cout << n << ' ';}
void co(void){cout << '\n';}
template <typename T>
void co(const T n){cout << n << '\n';}
template <typename T1, typename T2>
void co(pair<T1, T2> p){cout << p.fs << ' ' << p.sc << '\n';}
template <typename T>
void co(const Vec1<T> &v){for(T i : v) cosp(i); co();}
template <typename T>
void co(initializer_list<T> v){for(T i : v) cosp(i); co();}
template <typename T>
void ce(const T n){cerr << n << endl;}
void sonic(){ios::sync_with_stdio(false); cin.tie(0);}
void setp(const ll n){cout << fixed << setprecision(n);}
constexpr int INF = 1e9+1;
constexpr ll LINF = 1e18+1;
constexpr ll MOD = 1e9+7;
// constexpr ll MOD = 998244353;
constexpr ld EPS = 1e-11;
const ld PI = acos(-1);

struct rakuda{
	ll k, l, r;

	void input(){
		cin >> k >> l >> r;
	}
};

bool asc(rakuda &a, rakuda &b){
	return a.k < b.k;
}

void solve(){
	ll n;
	cin >> n;
	vector<rakuda> a(n);
	rep(i, n) a[i].input();

	sort(all(a), asc);

	priority_queue<ll> pq;
	ll ans = LINF;
	ll sum = 0;
	rep(i, n){
		if (a[i].l < a[i].r){
			sum += a[i].r;
			continue;
		}
		sum += a[i].l;
		pq.push(a[i].r - a[i].l);
		while(pq.size() > a[i].k){
			sum += pq.top();
			pq.pop();
		}
	}
	ce(sum);
	chmin(ans, sum);
	while (!pq.empty()) pq.pop();
	sum = 0;
	repr(i, n){
		if (a[i].l > a[i].r){
			sum += a[i].l;
			continue;
		}
		sum += a[i].r;
		pq.push(a[i].l - a[i].r);
		while(pq.size() > n - a[i].k){
			sum += pq.top();
			pq.pop();
		}
	}
	ce(sum);
	chmin(ans, sum);
	co(ans);
}

int main(void){
	ll t;
	cin >> t;
	rep(i, t) solve();

	return 0;
}