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;
const int N = 20005;
map<pair<int, int>, int> ch;
int n, m;
int totV = 1;
int S = 0;
int T = 1;
int tot = 1;
int head[N], np[N * 20], flow[N * 20], p[N * 20];
int A[55], B[55], C[55], l[55], r[55];
const int inf = 1e9;
const int Max = 1e7;
inline void link(int a, int b, int c) {
  tot++;
  p[tot] = b;
  np[tot] = head[a];
  head[a] = tot;
  flow[tot] = c;
  tot++;
  p[tot] = a;
  np[tot] = head[b];
  head[b] = tot;
  flow[tot] = 0;
}
int q[N], dis[N];
int BFS() {
  memset(dis, -1, sizeof dis);
  q[q[0] = 1] = S;
  dis[S] = 0;
  for (int i = (1); i <= (int)q[0]; i++) {
    int x = q[i];
    for (int u = head[x]; u; u = np[u])
      if (flow[u] && dis[p[u]] == -1) {
        dis[p[u]] = dis[x] + 1;
        q[++q[0]] = p[u];
      }
  }
  return dis[T];
}
int dinic(int x, int fl) {
  int ret = 0;
  if (x == T) return fl;
  for (int u = head[x]; u; u = np[u])
    if (flow[u] && dis[p[u]] == dis[x] + 1) {
      int tmp = dinic(p[u], min(fl, flow[u]));
      flow[u] -= tmp;
      flow[u ^ 1] += tmp;
      ret += tmp;
      fl -= tmp;
    }
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (1); i <= (int)n; i++) scanf("%d%d%d", &A[i], &B[i], &C[i]);
  for (int i = (1); i <= (int)n; i++) scanf("%d%d", &l[i], &r[i]);
  for (int i = (1); i <= (int)n; i++) {
    for (int j = (l[i] - 1); j <= (int)r[i]; j++) {
      pair<int, int> fuck = pair<int, int>(i, j);
      ch[fuck] = ++totV;
      if (j == l[i] - 1)
        link(S, totV, inf);
      else
        link(totV - 1, totV, Max - (A[i] * j * j + B[i] * j + C[i]));
    }
    link(totV, T, inf);
  }
  while (m--) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    for (int i = (l[u] - 1); i <= (int)r[u]; i++)
      if (i - d >= l[v] - 1 && i - d <= r[v]) {
        pair<int, int> f1 = pair<int, int>(u, i);
        pair<int, int> f2 = pair<int, int>(v, i - d);
        link(ch[f1], ch[f2], inf);
      }
  }
  int ans = 0;
  while (BFS() != -1) {
    ans += dinic(S, inf);
  }
  printf("%d\n", Max * n - ans);
  return 0;
}