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: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define NDEBUG
#include <cassert>


typedef long long ll;
typedef long double Double;
typedef unsigned long long ull;
typedef pair<int, int> ii;
typedef pair<ll, ll> llll;
typedef pair<double, double> dd;

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<llll> vllll;
typedef vector<bool> vb;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<long double> vD;

#define sz(a) int((a).size())
#define pb push_back
#define eb emplace_back
#define FOR(var, from, to) for (int var = (from); var <= (to); ++var)
#define rep(var, n) for (int var = 0; var < (n); ++var)
#define rep1(var, n) for (int var = 1; var <= (n); ++var)
#define repC2(vari, varj, n)               \
  for (int vari = 0; vari < (n)-1; ++vari) \
    for (int varj = vari + 1; varj < (n); ++varj)
#define repC3(vari, varj, vark, n)                  \
  for (int vari = 0; vari < (n)-2; ++vari)          \
    for (int varj = vari + 1; varj < (n)-1; ++varj) \
      for (int vark = varj + 1; vark < (n); ++vark)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define tr(i, c) for (auto i = (c).begin(); i != (c).end(); ++i)
#define found(s, e) ((s).find(e) != (s).end())
#define mset(arr, val) memset(arr, val, sizeof(arr))
#define mid(x, y) ((x) + ((y) - (x)) / 2)
#define IN(x, a, b) ((a) <= (x) && (x) <= (b))
#define cons make_pair
#define clamp(v, lo, hi) min(max(v, lo), hi)
#define ABS(x) max((x), -(x))
#define PQ(T) priority_queue<T, vector<T>, greater<T>>

template <typename T1, typename T2>
inline void amin(T1& a, T2 const& b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void amax(T1& a, T2 const& b) {
  if (a < b) a = b;
}
template <typename X, typename T>
auto vectors(X x, T a) {
  return vector<T>(x, a);
}
template <typename X, typename Y, typename Z, typename... Zs>
auto vectors(X x, Y y, Z z, Zs... zs) {
  auto cont = vectors(y, z, zs...);
  return vector<decltype(cont)>(x, cont);
}

inline ll square(ll x) { return x * x; }
inline ll gcd(ll a, ll b) {
  while (a) swap(a, b %= a);
  return b;
}
inline ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
template <typename T>
inline T mod(T a, T b) {
  return ((a % b) + b) % b;
}

template <typename T>
int find_left(vector<T>& v, T elem) {
  return (int)(upper_bound(v.begin(), v.end(), elem) - v.begin()) - 1;
}
template <typename T>
int find_right(vector<T>& v, T elem) {
  return (int)(lower_bound(v.begin(), v.end(), elem) - v.begin());
}

const ll MOD = 1000000007LL;

inline ll ADD(ll x, ll y) { return (x + y) % MOD; }
inline ll SUB(ll x, ll y) { return (x - y + MOD) % MOD; }
inline ll MUL(ll x, ll y) { return x * y % MOD; }
inline ll POW(ll x, ll e) {
  ll v = 1;
  for (; e; x = MUL(x, x), e >>= 1)
    if (e & 1) v = MUL(v, x);
  return v;
}
inline ll INV(ll y) { /*assert(y%MOD!=0);*/
  return POW(y, MOD - 2);
}
inline ll DIV(ll x, ll y) { return MUL(x, INV(y)); }

#define INTSPACE 12
char _buf[INTSPACE * 1000000 + 3];

int loadint() {
  if (fgets(_buf, INTSPACE + 3, stdin) == NULL) return 0;
  return atoi(_buf);
}

int loadvec(vector<int>& v, int N = -1) {
  if (N == 0) {
    v.clear();
    return 0;
  }
  if (N == -1) {
    N = loadint();
    if (N == 0) return 0;
  }
  int bufsize = INTSPACE * N + 3;
  if (fgets(_buf, bufsize, stdin) == NULL) return 0;
  v.resize(N);

  int i = 0;
  bool last = false;
  for (char* p = &_buf[0];;) {
    char* q = p;
    while (*q > ' ') ++q;
    if (*q == 0x0D || *q == 0x0A) last = true;
    *q = 0;
    v[i++] = atoi(p);
    if (last || i == N) break;
    p = q + 1;
  }
  return i;
}
void read_cr() { fgets(_buf, 256, stdin); }

void horizontal(vector<int>& v) {
  int L = v.size();
  for (int i = 0; i < L; ++i) {
    printf("%d%c", v[i], (i == L - 1) ? '\n' : ' ');
  }
}
void horizontall(vector<long long>& v) {
  int L = v.size();
  for (int i = 0; i < L; ++i) {
    printf("%lld%c", v[i], (i == L - 1) ? '\n' : ' ');
  }
}

ll sub(vii& a) {
  ll totalpen = 0;
  sort(ALL(a));
  int L = a.size();

  PQ(int) pq;
  int ent = 0;
  rep(i, L) {
    int w = a[i].first, pen = a[i].second;
    if (w >= 1 + ent) {
      ++ent;
      pq.push(pen);
    } else {
      if (!pq.empty() && pen > pq.top()) {
        totalpen += pq.top();
        pq.pop();

        pq.push(pen);
      } else {
        totalpen += pen;
      }
    }
  }
  return totalpen;
}

ll solve(int N, vi& k, vi& l, vi& r) {
  ll score = 0;
  vii left, right;
  rep(i, N) {
    if (k[i] == N || l[i] == r[i]) {
      score += l[i];
      continue;
    }

    if (l[i] > r[i]) {
      left.eb(k[i], l[i] - r[i]);
      score += l[i];
    } else {
      right.eb(N - k[i], r[i] - l[i]);
      score += r[i];
    }
  }
  return score - sub(left) - sub(right);
}

int main() {
  int T;
  scanf("%d%*c", &T);
  rep(t, T) {
    int N;
    scanf("%d%*c", &N);
    vi k(N), l(N), r(N);
    rep(i, N) {
      scanf("%d %d %d%*c", &k[i], &l[i], &r[i]);
    }
    ll ans = solve(N, k, l, r);
    printf("%lld\n", ans);
  }

  return 0;
}