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 inf = 1e9;
int id[110][210];
int n, m, S, T;
int ans;
int l[110], r[110], a[110], b[110], c[110];
int d[200010];
int tot, num;
int cur[200010], head[200010];
int vel[2000010], key[2000010], ne[2000010], addnum;
void add(int u, int v, int k) {
  addnum++, vel[addnum] = v, key[addnum] = k, ne[addnum] = head[u],
            head[u] = addnum;
  addnum++, vel[addnum] = u, key[addnum] = 0, ne[addnum] = head[v],
            head[v] = addnum;
}
bool bfs() {
  queue<int> q;
  for (int i = 0; i <= tot; i++) d[i] = -1;
  d[S] = 0;
  q.push(S);
  while (q.size()) {
    int u = q.front();
    q.pop();
    for (int e = head[u]; e; e = ne[e]) {
      if (!key[e]) continue;
      int v = vel[e];
      if (d[v] == -1) {
        d[v] = d[u] + 1;
        q.push(v);
      }
    }
  }
  return d[T] != -1;
}
int dfs(int u, int k) {
  if (u == T) return k;
  int used = 0;
  for (int &e = cur[u]; e; e = ne[e]) {
    if (!key[e]) continue;
    int v = vel[e];
    if (d[v] == d[u] + 1) {
      int t = dfs(v, min(k - used, key[e]));
      key[e] -= t;
      key[e ^ 1] += t;
      used += t;
      if (used == k) return k;
    }
  }
  return used;
}
int main() {
  addnum = 1;
  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]);
  tot = 1;
  S = 0;
  T = 1;
  num = 1e6;
  for (int i = 1; i <= n; i++) {
    for (int j = l[i]; j <= r[i]; j++) {
      id[i][j - l[i]] = ++tot;
      add(tot, tot + 1, num - (a[i] * j * j + b[i] * j + c[i]));
    }
    add(S, id[i][0], inf);
    id[i][r[i] - l[i] + 1] = ++tot;
    add(tot, T, inf);
  }
  for (int i = 1; i <= m; i++) {
    int x, y, d;
    scanf("%d%d%d", &x, &y, &d);
    for (int j = max(l[x] - d, l[y]); j <= min(r[x] - d, r[y]) + 1; j++)
      add(id[x][j + d - l[x]], id[y][j - l[y]], inf);
  }
  while (bfs()) {
    for (int i = 0; i <= tot; i++) cur[i] = head[i];
    ans += dfs(S, inf);
  }
  printf("%d\n", num * n - ans);
}