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;
struct edge {
  int f, t;
  long long cap;
} E[2000005];
vector<int> adj[20005];
queue<int> q;
int path[20005];
int S, T, N;
void add_edge(int u, int v, long long cap) {
  E[N].f = u, E[N].t = v, E[N].cap = cap;
  adj[u].push_back(N), N++;
  E[N].f = v, E[N].t = u, E[N].cap = 0;
  adj[v].push_back(N), N++;
}
bool find() {
  while (q.size()) q.pop();
  memset(path, -1, sizeof(path));
  path[S] = -2, q.push(S);
  while (q.size()) {
    int u = q.front();
    q.pop();
    if (u == T) return 1;
    for (int i = 0; i < adj[u].size(); ++i) {
      int v = E[adj[u][i]].t;
      if (E[adj[u][i]].cap > 0 && path[v] == -1) path[v] = adj[u][i], q.push(v);
    }
  }
  return 0;
}
long long flow() {
  long long ret = 0;
  while (find()) {
    long long bot = 1ll << 60;
    int u = T;
    while (path[u] >= 0) {
      bot = min(bot, E[path[u]].cap);
      u = E[path[u]].f;
    }
    u = T;
    while (path[u] >= 0) {
      E[path[u]].cap -= bot;
      E[path[u] ^ 1].cap += bot;
      u = E[path[u]].f;
    }
    ret += bot;
  }
  return ret;
}
int n, m;
int a[55], b[55], c[55];
int L[55], R[55];
int pad = 110;
int node(int u, int x) { return u * pad * 2 + x + pad; }
int main() {
  S = 18000, T = 19000;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) scanf("%d%d%d", a + i, b + i, c + i);
  for (int i = 0; i < n; ++i) scanf("%d%d", L + i, R + i);
  for (int u = 0; u < n; ++u) {
    add_edge(S, node(u, L[u] - 1), (1ll << 50));
    for (int x = L[u] - 1; x < R[u]; ++x)
      add_edge(
          node(u, x), node(u, x + 1),
          (1ll << 20) - (a[u] * (x + 1) * (x + 1) + b[u] * (x + 1) + c[u]));
    add_edge(node(u, R[u]), T, (1ll << 50));
  }
  for (int z = 0; z < m; ++z) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d), u--, v--;
    for (int x = L[u] - 1; x <= R[u]; ++x)
      if (L[v] - 1 <= x - d && x - d <= R[v])
        add_edge(node(u, x), node(v, x - d), (1ll << 50));
  }
  long long f = flow();
  cout << (1ll << 20) * n - f << endl;
}