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 T = 20001;
const long long oo = 10000000000ll;
struct Edge {
  int y;
  long long v;
  Edge *next, *op;
};
Edge *b[20010], *cur[20010];
int h[20010];
long long ans;
void addedge(int x, int y, long long z) {
  Edge *p = new Edge;
  p->y = y;
  p->v = z;
  p->next = b[x];
  b[x] = p;
  Edge *p1 = p;
  p = new Edge;
  p->y = x;
  p->v = 0;
  p->next = b[y];
  b[y] = p;
  p1->op = p;
  p->op = p1;
}
bool bfs() {
  queue<int> q;
  memset(h, -1, sizeof(h));
  h[0] = 0;
  q.push(0);
  for (; !q.empty();) {
    int x = q.front();
    q.pop();
    for (Edge *i = b[x]; i != NULL; i = i->next)
      if (i->v && h[i->y] == -1) {
        h[i->y] = h[x] + 1;
        q.push(i->y);
      }
  }
  return h[T] != -1;
}
long long dfs(int x, long long f) {
  if (x == T) return f;
  long long w, used = 0;
  for (Edge *i = cur[x]; i != NULL; i = i->next)
    if (h[x] + 1 == h[i->y]) {
      w = f - used;
      w = dfs(i->y, min(i->v, w));
      i->v -= w;
      i->op->v += w;
      if (i->v) cur[x] = i;
      used += w;
      if (used == f) return f;
    }
  if (!used) h[x] = -1;
  return used;
}
void dinic() {
  for (; bfs();) {
    for (int i = 0; i <= 20001; i++) cur[i] = b[i];
    ans += dfs(0, oo);
  }
}
int A[51], B[51], C[51], l[51], r[51], st[51];
int f(int i, int x) { return A[i] * x * x + B[i] * x + C[i]; }
int node(int x, int y) { return st[x] + y - l[x]; }
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d%d%d", &A[i], &B[i], &C[i]);
  int tot = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    st[i] = tot;
    tot += r[i] - l[i] + 2;
  }
  int mx = 210000;
  for (int i = 1; i <= n; i++) {
    addedge(0, st[i], oo);
    for (int j = l[i]; j <= r[i]; j++)
      addedge(node(i, j), node(i, j) + 1, mx - f(i, j));
    addedge(node(i, r[i]) + 1, T, oo);
  }
  for (int i = 0; i < m; i++) {
    int x, y, v;
    scanf("%d%d%d", &x, &y, &v);
    for (int j = l[x]; j <= r[x]; j++)
      if (j - v >= l[y] && j - v <= r[y] + 1)
        addedge(node(x, j), node(y, j - v), oo);
  }
  dinic();
  printf("%I64d\n", (long long)mx * n - ans);
  return 0;
}