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;
int n, m;
int a[55], b[55], c[55], lim[55][2], id[55][210], num;
int S, T, MAX;
inline int f(int i, int j) { return a[i] * j * j + b[i] * j + c[i]; }
int tot = 1, there[80000], nx[80000], fl[80000], head[12000];
inline void addedge(int x, int y, int flow) {
  ++tot;
  there[tot] = y;
  nx[tot] = head[x];
  head[x] = tot;
  fl[tot] = flow;
  ++tot;
  there[tot] = x;
  nx[tot] = head[y];
  head[y] = tot;
  fl[tot] = 0;
}
int d[12000], v[12000];
int isap(int x, int flow) {
  if (x == T) return flow;
  int now = 0;
  for (int i = head[x]; i; i = nx[i])
    if (fl[i] > 0 && d[x] == d[there[i]] + 1) {
      int tt = isap(there[i], min(flow - now, fl[i]));
      fl[i] -= tt;
      fl[i ^ 1] += tt;
      now += tt;
      if (now == flow) return now;
      if (d[S] >= T) return now;
    }
  if (!(--v[d[x]])) d[S] = T;
  v[++d[x]]++;
  return now;
}
int main() {
  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", &lim[i][0], &lim[i][1]);
  for (int i = 1; i <= n; ++i)
    for (int j = lim[i][0]; j <= lim[i][1] + 1; ++j) {
      if (j != lim[i][1] + 1) MAX = max(MAX, f(i, j));
      id[i][j - lim[i][0]] = ++num;
    }
  S = ++num, T = ++num;
  for (int i = 1; i <= n; ++i) {
    addedge(S, id[i][0], 1000000000);
    addedge(id[i][lim[i][1] + 1 - lim[i][0]], T, 1000000000);
    for (int j = lim[i][0]; j <= lim[i][1]; ++j)
      addedge(id[i][j - lim[i][0]], id[i][j + 1 - lim[i][0]], MAX - f(i, j));
  }
  int _u, _v, dd;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &_u, &_v, &dd);
    for (int j = max(lim[_u][0], lim[_v][0] + dd) - 1;
         j <= min(lim[_u][1], lim[_v][1] + dd); ++j) {
      addedge(id[_u][j - lim[_u][0] + 1], id[_v][j - dd - lim[_v][0] + 1],
              1000000000);
    }
  }
  int ans = 0;
  v[0] = T;
  while (d[S] < T) ans += isap(S, 1000000000);
  printf("%d\n", MAX * n - ans);
  return 0;
}