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[23333], b[23333], c[23333], l[23333], r[23333], s[23333], H[23333];
const int inf = 2147483647;
inline int f(int i, int x) { return x * a[i] * x + b[i] * x + c[i]; }
inline int get(int i, int j) { return s[i] + j - l[i]; }
struct T {
  int from, to, nxt, v;
} way[1 << 21];
int h[23333], num = 1;
int dis[23333];
inline void adde(int a, int b, int f) {
  way[++num] = (T){a, b, h[a], f}, h[a] = num;
  way[++num] = (T){b, a, h[b], 0}, h[b] = num;
}
queue<int> q;
int T;
inline int bfs(int s, int t) {
  for (int i = 0; i <= T; ++i) dis[i] = inf, H[i] = h[i];
  dis[s] = 0;
  q.push(s);
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    for (int i = h[t]; i; i = way[i].nxt)
      if (dis[way[i].to] == inf && way[i].v)
        dis[way[i].to] = dis[t] + 1, q.push(way[i].to);
  }
  return dis[t] != inf;
}
inline int dfs(int s, int to, int f) {
  if (s == to || f == 0) return f;
  int ans = 0, m;
  for (int& i = H[s]; i; i = way[i].nxt)
    if (dis[way[i].to] > dis[s] && (m = dfs(way[i].to, to, min(f, way[i].v)))) {
      ans += m;
      way[i ^ 1].v += m;
      way[i].v -= m;
      f -= m;
      if (f == 0) break;
    }
  return ans;
}
inline int dinic(int s, int t) {
  int ans = 0;
  while (bfs(s, t)) ans += dfs(s, t, inf);
  return ans;
}
int main() {
  scanf("%d", &(n)), scanf("%d", &(m));
  for (int i = 1; i <= n; ++i)
    scanf("%d", &(a[i])), scanf("%d", &(b[i])), scanf("%d", &(c[i]));
  int t = 1, MAXN = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &(l[i])), scanf("%d", &(r[i])), s[i] = t, t += r[i] - l[i] + 2;
    for (int j = l[i]; j <= r[i]; ++j) MAXN = max(MAXN, f(i, j));
  }
  for (int i = 1; i <= n; ++i) {
    adde(0, s[i], inf);
    for (int j = l[i]; j <= r[i]; ++j)
      adde(get(i, j), get(i, j) + 1, MAXN - f(i, j));
    adde(get(i, r[i]) + 1, t, inf);
  }
  for (int i = 1, x, y, z; i <= m; ++i) {
    scanf("%d", &(x)), scanf("%d", &(y)), scanf("%d", &(z));
    for (int j = l[x]; j <= r[x]; ++j)
      if (j - z >= l[y] && j - z <= r[y] + 1)
        adde(get(x, j), get(y, j - z), inf);
  }
  cout << n * MAXN - dinic(0, T = t);
}