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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, l[50 + 5], r[50 + 5], A[50 + 5], B[50 + 5], C[50 + 5], ans, en,
    cnt = 1;
int c[10051 + 5], d[10051 + 5], head[10051 + 5], q[10051 + 5], top,
    id[50 + 5][205];
struct edge {
  int to, next, w;
} e[2000005];
inline void ins(int f, int t, int w) {
  e[++cnt] = (edge){t, head[f], w};
  head[f] = cnt;
  e[++cnt] = (edge){f, head[t], 0};
  head[t] = cnt;
}
bool bfs() {
  memset(d, 0, sizeof(d));
  int i, j;
  for (d[q[top = i = 1] = 0] = 1; i <= top; ++i)
    for (j = c[q[i]] = head[q[i]]; j; j = e[j].next)
      if (e[j].w && !d[e[j].to]) d[q[++top] = e[j].to] = d[q[i]] + 1;
  return d[10051];
}
int dfs(int x, int f) {
  if (x == 10051) return f;
  int used = 0;
  for (int& i = c[x]; i; i = e[i].next)
    if (e[i].w && d[e[i].to] == d[x] + 1) {
      int w = dfs(e[i].to, min(f - used, e[i].w));
      e[i].w -= w;
      e[i ^ 1].w += w;
      used += w;
      if (used == f) return f;
    }
  return d[x] = -1, used;
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= n; ++i) A[i] = read(), B[i] = read(), C[i] = read();
  for (int i = 1; i <= n; ++i) {
    l[i] = read(), r[i] = read();
    int mx = -1e9;
    for (int j = l[i]; j <= r[i]; ++j)
      mx = max(mx, A[i] * j * j + B[i] * j + C[i]), id[i][j + 100] = ++en;
    ans += mx;
    ins(0, id[i][l[i] + 100], 2000000000);
    for (int j = l[i]; j <= r[i]; ++j)
      ins(id[i][j + 100], j < r[i] ? id[i][j + 101] : 10051,
          mx - (A[i] * j * j + B[i] * j + C[i]));
  }
  for (int i = 1; i <= m; ++i) {
    int u = read(), v = read(), d = read();
    for (int j = l[u]; j <= r[u]; ++j)
      if (j - d > r[v])
        ins(id[u][j + 100], 10051, 2000000000);
      else if (j - d >= l[v])
        ins(id[u][j + 100], id[v][j - d + 100], 2000000000);
  }
  while (bfs()) ans -= dfs(0, 2000000000);
  cout << ans;
  return 0;
}