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>
struct Edge {
  int v, x, cap;
} E[2222222];
int l[10100], tot;
const int inf = 1e9 + 7;
void add_edge(int u, int v, int cap) {
  E[tot].v = v;
  E[tot].cap = cap;
  E[tot].x = l[u];
  l[u] = tot++;
  E[tot].v = u;
  E[tot].cap = 0;
  E[tot].x = l[v];
  l[v] = tot++;
}
int d[10100], queue[10100];
bool bfs(int src, int sink) {
  int s = 0, t = 0;
  memset(d, -1, sizeof d);
  d[src] = 0;
  queue[t++] = src;
  while (s < t) {
    int u = queue[s++];
    for (int i = l[u]; ~i; i = E[i].x) {
      int v = E[i].v;
      if (d[v] == -1 && E[i].cap) {
        d[queue[t++] = v] = d[u] + 1;
      }
    }
  }
  return d[sink] != -1;
}
int dfs(int u, int in, int sink) {
  if (u == sink) return in;
  int flow = 0, f;
  for (int i = l[u]; ~i && flow < in; i = E[i].x) {
    int v = E[i].v;
    if (d[v] == d[u] + 1 && E[i].cap) {
      if (f = dfs(v, std::min(E[i].cap, in - flow), sink)) {
        E[i].cap -= f;
        E[i ^ 1].cap += f;
        flow += f;
      }
    }
  }
  if (flow < in) d[u] = -1;
  return flow;
}
int Dinic(int src, int sink) {
  int flow = 0;
  while (bfs(src, sink)) flow += dfs(src, inf, sink);
  return flow;
}
int a[50], b[50], c[50], down[50], up[50];
std::map<std::pair<int, int>, int> id;
int calc(int i, int x) { return a[i] * x * x + b[i] * x + c[i]; }
int main() {
  int N, M, u, v, d, n = 0;
  int src, sink;
  std::memset(l, -1, sizeof l);
  tot = 0;
  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", &down[i], &up[i]);
    for (int x = down[i] - 1; x <= up[i]; x++) {
      id[std::make_pair(i, x)] = ++n;
    }
  }
  src = ++n;
  sink = ++n;
  for (int i = 0; i < N; i++) {
    for (int x = down[i] - 1; x < up[i]; x++) {
      u = id[std::make_pair(i, x)];
      add_edge(u, u + 1, 10000000 - calc(i, x + 1));
    }
    add_edge(src, id[std::make_pair(i, down[i] - 1)], inf);
    add_edge(id[std::make_pair(i, up[i])], sink, inf);
  }
  for (int i = 0; i < M; i++) {
    scanf("%d%d%d", &u, &v, &d);
    u--, v--;
    for (int x = down[u] - 1; x < up[u]; x++) {
      if (!id.count(std::make_pair(v, x - d))) continue;
      add_edge(id[std::make_pair(u, x)], id[std::make_pair(v, x - d)], inf);
    }
  }
  int ans = 10000000 * N - Dinic(src, sink);
  printf("%d\n", ans);
  return 0;
}