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;
namespace Flow {
int S = 1, T = 2, D = 1, C = 2, Q[200005], st[200005], Dis[200005];
struct Lin {
  int v, f, next;
} E[200005];
void Link(int u, int v, int f) {
  E[++D] = (Lin){v, f, st[u]};
  st[u] = D;
  E[++D] = (Lin){u, 0, st[v]};
  st[v] = D;
}
bool BFS() {
  for (int i = 1; i <= C; i++) Dis[i] = -1;
  Dis[Q[1] = S] = 0;
  for (int l = 0, r = 1; l < r;) {
    int u = Q[++l];
    for (int i = st[u], v; i; i = E[i].next)
      if (E[i].f && Dis[v = E[i].v] == -1) {
        Dis[Q[++r] = v] = Dis[u] + 1;
        if (v == T) return 1;
      }
  }
  return 0;
}
int DFS(int u, int Lim) {
  if (u == T) return Lim;
  int c = 0;
  for (int i = st[u], v; i; i = E[i].next)
    if (E[i].f && Dis[u] + 1 == Dis[v = E[i].v]) {
      int t = DFS(v, ((E[i].f) < (Lim - c) ? (E[i].f) : (Lim - c)));
      if (t) {
        E[i].f -= t;
        E[i ^ 1].f += t;
        c += t;
        if (c == Lim) break;
      } else
        Dis[v] = -1;
    }
  return c;
}
int Max_Flow() {
  int ret = 0;
  for (; BFS();) ret += DFS(S, (1 << 30));
  return ret;
}
};  // namespace Flow
int N, M, A[55], B[55], C[55], L[55], R[55], G[55];
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 1; i <= N; i++) scanf("%d%d%d", &A[i], &B[i], &C[i]);
  for (int i = 1; i <= N; i++) {
    scanf("%d%d", &L[i], &R[i]);
    Flow::Link(Flow::S, ++Flow::C, (1 << 30));
    G[i] = Flow::C;
    for (int j = L[i]; j <= R[i]; j++)
      Flow::Link(Flow::C, Flow::C + 1,
                 1000000 - (A[i] * j * j + B[i] * j + C[i])),
          ++Flow::C;
    Flow::Link(Flow::C, Flow::T, (1 << 30));
  }
  for (int i = 1; i <= M; i++) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    for (int j = L[u]; j <= R[u]; j++)
      if (j - d >= L[v] && j - d <= R[v])
        Flow::Link(G[u] + j - L[u], G[v] + j - d - L[v], (1 << 30));
      else if (j - d > R[v])
        Flow::Link(G[u] + j - L[u], Flow::T, (1 << 30));
  }
  printf("%d\n", N * 1000000 - Flow::Max_Flow());
}