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 = 10505;
int n, Q, id[55][205], S, T;
int head[N], tot = 1, nd;
int dis[N], q[N], ans;
struct edge {
  int to, next, f;
} e[N * 20];
void add(int x, int y, int f) {
  e[++tot] = (edge){y, head[x], f};
  head[x] = tot;
  e[++tot] = (edge){x, head[y], 0};
  head[y] = tot;
}
bool bfs() {
  for (int i = (int)(1); i <= (int)(nd); i++) dis[i] = -1;
  int h = 0, t = 1;
  q[1] = S;
  dis[S] = 0;
  while (h != t) {
    int x = q[++h];
    for (int i = head[x]; i; i = e[i].next)
      if (dis[e[i].to] == -1 && e[i].f) {
        dis[e[i].to] = dis[x] + 1;
        if (e[i].to == T) return 1;
        q[++t] = e[i].to;
      }
  }
  return 0;
}
int dfs(int x, int flow) {
  if (x == T) return flow;
  int k, rest = flow;
  for (int i = head[x]; i && rest; i = e[i].next)
    if (dis[e[i].to] == dis[x] + 1 && e[i].f) {
      k = dfs(e[i].to, min(rest, e[i].f));
      e[i].f -= k;
      e[i ^ 1].f += k;
      rest -= k;
    }
  if (rest) dis[x] = -1;
  return flow - rest;
}
int main() {
  scanf("%d%d", &n, &Q);
  S = ++nd;
  T = ++nd;
  for (int i = (int)(1); i <= (int)(n); i++)
    for (int j = (int)(0); j <= (int)(201); j++) id[i][j] = ++nd;
  for (int i = (int)(1); i <= (int)(n); i++) {
    int A, B, C;
    scanf("%d%d%d", &A, &B, &C);
    for (int j = (int)(-100); j <= (int)(100); j++)
      add(id[i][j + 100], id[i][j + 101], 1000000 - A * j * j - B * j - C);
  }
  for (int i = (int)(1); i <= (int)(n); i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    add(S, id[i][l + 100], 1 << 30);
    add(id[i][r + 101], T, 1 << 30);
  }
  for (int i = (int)(1); i <= (int)(Q); i++) {
    int x, y, v;
    scanf("%d%d%d", &y, &x, &v);
    for (int j = (int)(-100); j <= (int)(100); j++)
      add(id[y][j + 100], id[x][max(min(j + 100 - v, 201), 0)], 1 << 30);
  }
  for (; bfs(); ans += dfs(S, 1 << 30))
    ;
  printf("%d\n", 1000000 * n - ans);
}