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;
template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++i) {
    if (i != v.begin()) os << ", ";
    os << *i;
  }
  os << "]";
  return os;
}
int n, m;
vector<int> A, B, C, L, R;
inline int get_value(int i, int x) { return A[i] * x * x + B[i] * x + C[i]; }
const int source = 0;
const int sink = 1;
int num_nodes = 2;
vector<map<int, int> > nodes;
vector<vector<int> > G;
int main() {
  scanf("%d%d", &n, &m);
  A.resize(n);
  B.resize(n);
  C.resize(n);
  L.resize(n);
  R.resize(n);
  for (int i = 0; i < (int)(n); ++i) scanf("%d%d%d", &A[i], &B[i], &C[i]);
  int min_value = 1023456789, max_value = -1023456789;
  for (int i = 0; i < (int)(n); ++i) {
    scanf("%d%d", &L[i], &R[i]);
    for (int j = (int)(L[i]); j <= (int)(R[i]); ++j) {
      int v = get_value(i, j);
      min_value = min(min_value, v);
      max_value = max(max_value, v);
    }
  }
  int bound = (n + 1) * max_value - n * min_value + 1;
  nodes.resize(n);
  for (int i = 0; i < (int)(n); ++i)
    for (int j = (int)(L[i] - 1); j <= (int)(R[i]); ++j)
      nodes[i][j] = num_nodes++;
  G.resize(num_nodes, vector<int>(num_nodes, 0));
  for (int i = 0; i < (int)(n); ++i) {
    for (int j = (int)(L[i]); j <= (int)(R[i]); ++j)
      G[nodes[i][j - 1]][nodes[i][j]] = bound - get_value(i, j);
    G[source][nodes[i][L[i] - 1]] = 1023456789;
    G[nodes[i][R[i]]][sink] = 1023456789;
  }
  for (int i = 0; i < (int)(m); ++i) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    --u, --v;
    for (int j = (int)(L[u] - 1); j <= (int)(R[u]); ++j)
      if (L[v] <= j - d && j - d <= R[v])
        G[nodes[u][j]][nodes[v][j - d]] = 1023456789;
  }
  int flow = 0;
  for (;;) {
    vector<int> P(num_nodes, -1);
    P[source] = -2;
    vector<int> F(num_nodes, 0);
    F[source] = 1023456789;
    queue<int> Q;
    Q.push(source);
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (int i = 0; i < (int)(num_nodes); ++i)
        if (G[x][i] && P[i] == -1) {
          P[i] = x;
          F[i] = min(F[x], G[x][i]);
          Q.push(i);
        }
      if (P[sink] != -1) break;
    }
    if (F[sink] == 0) break;
    int x = sink;
    while (P[x] != -2) {
      G[P[x]][x] -= F[sink];
      G[x][P[x]] += F[sink];
      x = P[x];
    }
    flow += F[sink];
  }
  printf("%d\n", n * bound - flow);
}