Problem: Nanami likes playing games, and is also really good at it. This day she was playing a new game which involved operating a power plant. Nanami's job is to control the generators in the plant and produce maximum output.

There are n generators in the plant. Each generator should be set to a generating level. Generating level is an integer (possibly zero or negative), the generating level of the i-th generator should be between li and ri (both inclusive). The output of a generator can be calculated using a certain quadratic function f(x), where x is the generating level of the generator. Each generator has its own function, the function of the i-th generator is denoted as fi(x).

However, there are m further restrictions to the generators. Let the generating level of the i-th generator be xi. Each restriction is of the form xu ≤ xv + d, where u and v are IDs of two different generators and d is an integer.

Nanami found the game tedious but giving up is against her creed. So she decided to have a program written to calculate the answer for her (the maximum total output of generators). Somehow, this became your job.

Input

The first line contains two integers n and m (1 ≤ n ≤ 50; 0 ≤ m ≤ 100) — the number of generators and the number of restrictions.

Then follow n lines, each line contains three integers ai, bi, and ci (|ai| ≤ 10; |bi|, |ci| ≤ 1000) — the coefficients of the function fi(x). That is, fi(x) = aix2 + bix + ci.

Then follow another n lines, each line contains two integers li and ri ( - 100 ≤ li ≤ ri ≤ 100).

Then follow m lines, each line contains three integers ui, vi, and di (1 ≤ ui, vi ≤ n; ui ≠ vi; |di| ≤ 200), describing a restriction. The i-th restriction is xui ≤ xvi + di.

Output

Print a single line containing a single integer — the maximum output of all the generators. It is guaranteed that there exists at least one valid configuration.

Examples

Input

3 3
0 1 0
0 1 1
0 1 2
0 3
1 2
-100 100
1 2 0
2 3 0
3 1 0


Output

9


Input

5 8
1 -8 20
2 -4 0
-1 10 -10
0 1 0
0 -1 1
1 9
1 4
0 10
3 11
7 9
2 1 3
1 2 3
2 3 3
3 2 3
3 4 3
4 3 3
4 5 3
5 4 3


Output

46

Note

In the first sample, f1(x) = x, f2(x) = x + 1, and f3(x) = x + 2, so we are to maximize the sum of the generating levels. The restrictions are x1 ≤ x2, x2 ≤ x3, and x3 ≤ x1, which gives us x1 = x2 = x3. The optimal configuration is x1 = x2 = x3 = 2, which produces an output of 9.

In the second sample, restrictions are equal to |xi - xi + 1| ≤ 3 for 1 ≤ i < n. One of the optimal configurations is x1 = 1, x2 = 4, x3 = 5, x4 = 8 and x5 = 7.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const pair<int, int> DD[] = {{1, 0},  {1, 1},   {0, 1},  {-1, 1},
                             {-1, 0}, {-1, -1}, {0, -1}, {1, -1}};
int N;
int F[1 << 20];
vector<pair<int, int> > s[1 << 20];
int m = -1;
void add(int a, int b, int c) {
  F[++m] = c;
  s[a].push_back(make_pair(b, m));
  F[++m] = 0;
  s[b].push_back(make_pair(a, m));
}
long long bfs(void) {
  static int D[1 << 20];
  static int C[1 << 20];
  for (int i = 1; i <= N; ++i) D[i] = 0, C[i] = 0;
  D[1] = 1;
  queue<int> Q;
  Q.push(1);
  while (!Q.empty()) {
    int node = Q.front();
    Q.pop();
    for (auto it : s[node])
      if (F[it.second] > 0 && !D[it.first])
        D[it.first] = node, C[it.first] = it.second, Q.push(it.first);
  }
  if (!D[N]) return -1;
  int flow = 1 << 30;
  for (int i = N; i != 1; i = D[i]) flow = min(flow, F[C[i]]);
  for (int i = N; i != 1; i = D[i]) F[C[i]] -= flow, F[C[i] ^ 1] += flow;
  return flow;
}
int main(void) {
  srand(time(0));
  cout << fixed << setprecision(7);
  cerr << fixed << setprecision(7);
  int n, m;
  cin >> n >> m;
  static int a[1 << 20];
  static int b[1 << 20];
  static int c[1 << 20];
  auto f = [&](int i, int first) {
    return first * first * a[i] + first * b[i] + c[i];
  };
  for (int i = 1; i <= n; ++i) cin >> a[i] >> b[i] >> c[i];
  vector<pair<int, int> > edges(n + 5);
  for (int i = 1; i <= n; ++i) cin >> edges[i].first >> edges[i].second;
  ++N;
  static int D[555][555];
  int MAX = -(1 << 30);
  for (int i = 1; i <= n; ++i)
    for (int j = edges[i].first; j <= edges[i].second; ++j)
      MAX = max(MAX, f(i, j));
  for (int i = 1; i <= n; ++i) {
    add(1, ++N, 1 << 30);
    D[i][edges[i].first + 150] = N;
    for (int j = edges[i].first; j <= edges[i].second; ++j) {
      D[i][j + 151] = ++N;
      add(N - 1, N, MAX - f(i, j));
    }
  }
  while (m--) {
    int l, r, v;
    cin >> l >> r >> v;
    for (int j = edges[l].first; j <= edges[l].second; ++j)
      if (edges[r].first <= j - v && j - v <= edges[r].second + 1)
        add(D[l][j + 150], D[r][j - v + 150], 1 << 30);
  }
  ++N;
  for (int i = 1; i <= n; ++i) add(D[i][edges[i].second + 151], N, 1 << 30);
  long long flow = 0;
  do {
    long long cnt = bfs();
    if (cnt <= 0) break;
    flow += cnt;
  } while (1);
  cout << 1ll * MAX * n - flow << '\n';
  cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms"
       << '\n';
  return 0;
}