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 = 1 << 30;
const int MAXN = 10005;
const int MAXM = 300005;
int n, m, cnt, T, in[MAXN], out[MAXN], cur[MAXN], d[MAXN];
vector<int> Eid[MAXN];
struct Edge {
  int c, v;
} E[MAXM];
void AddEdge(int x, int y, int c) {
  E[++cnt].v = y, E[cnt].c = c;
  Eid[x].push_back(cnt);
  E[++cnt].v = x, E[cnt].c = 0;
  Eid[y].push_back(cnt);
}
bool BFS() {
  static int Q[MAXN];
  int x, i, ql, qr;
  for (i = 0; i <= T; i++) d[i] = -1;
  Q[ql = qr = 0] = 0;
  d[0] = 0;
  while (ql <= qr) {
    x = Q[ql++];
    for (i = 0; i < Eid[x].size(); i++) {
      const Edge &e = E[Eid[x][i]];
      if (e.c && d[e.v] == -1) {
        Q[++qr] = e.v;
        d[e.v] = d[x] + 1;
        if (e.v == T) return 1;
      }
    }
  }
  return 0;
}
int Aug(int x, int a) {
  if (x == T) return a;
  int aug = 0, delta;
  for (int &i = cur[x]; i < Eid[x].size(); i++) {
    Edge &e = E[Eid[x][i]];
    if (e.c && d[e.v] == d[x] + 1) {
      delta = Aug(e.v, min(a, e.c));
      if (delta) {
        e.c -= delta;
        E[Eid[x][i] ^ 1].c += delta;
        aug += delta;
        if (!(a -= delta)) break;
      }
    }
  }
  if (!aug) d[x] = -1;
  return aug;
}
int MaxFlow() {
  int flow = 0;
  while (BFS()) {
    for (int i = 0; i <= T; i++) cur[i] = 0;
    flow += Aug(0, inf);
  }
  return flow;
}
int a[MAXN], b[MAXN], c[MAXN], l[MAXN], r[MAXN], id[55][250], ans, tot;
inline int F(int i, int j) { return j * j * a[i] + j * b[i] + c[i]; }
int main() {
  cnt = -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);
    int tmp = 0;
    for (int j = l[i]; j <= r[i]; j++) tmp = max(tmp, F(i, j));
    ans += tmp;
    for (int j = l[i] - 1; j <= r[i]; j++) id[i][j - l[i] + 1] = ++tot;
    for (int j = l[i]; j <= r[i]; j++)
      AddEdge(id[i][j - l[i]], id[i][j - l[i] + 1], tmp - F(i, j));
  }
  while (m--) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    for (int i = l[u]; i <= r[u]; i++) {
      int j = i - d;
      if (j > r[v]) j = r[v] + 1;
      if (j >= l[v]) AddEdge(id[u][i - l[u]], id[v][j - l[v]], inf);
    }
  }
  T = ++tot;
  for (int i = 1; i <= n; i++)
    AddEdge(0, id[i][0], inf), AddEdge(id[i][r[i] - l[i] + 1], T, inf);
  ans -= MaxFlow();
  cout << ans;
}