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 = 4e6;
struct edge {
  int next, to;
  long long w;
} a[N];
int head[N], cnt = 1, n, m, cur[N], dep[N], S, T;
int L[N], R[N], nod;
queue<int> Q;
map<int, int> Map[51];
long long Ans, inf = 1e14;
struct function {
  long long a, b, c;
  void rev() {
    a = -a;
    b = -b;
    c = -c;
  }
  long long f(long long x) { return (1ll << 25) + (a * x * x + b * x + c); }
} F[N];
void link(int x, int y, long long w) {
  a[++cnt] = (edge){head[x], y, w};
  head[x] = cnt;
  a[++cnt] = (edge){head[y], x, 0};
  head[y] = cnt;
}
int BFS() {
  memset(dep, 0, sizeof(dep));
  Q.push(S);
  dep[S] = 1;
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = head[x]; i; i = a[i].next)
      if (!dep[a[i].to] && a[i].w) dep[a[i].to] = dep[x] + 1, Q.push(a[i].to);
  }
  return dep[T];
}
int DFS(int x, long long flow) {
  if (x == T) return flow;
  for (int &i = cur[x]; i; i = a[i].next) {
    int R = a[i].to;
    if (!a[i].w || dep[R] != dep[x] + 1) continue;
    long long k = DFS(R, min(flow, a[i].w));
    if (k) {
      a[i].w -= k, a[i ^ 1].w += k;
      return k;
    }
  }
  return 0;
}
int id(int p, int c) {
  int &v = Map[p][c];
  if (!v) v = ++nod;
  return v;
}
int main() {
  cin >> n >> m;
  S = ++nod;
  T = ++nod;
  for (int i = 1; i <= n; i++) cin >> F[i].a >> F[i].b >> F[i].c, F[i].rev();
  for (int i = 1; i <= n; i++) {
    cin >> L[i] >> R[i];
    link(S, id(i, L[i]), inf);
    for (int c = L[i]; c <= R[i]; c++) link(id(i, c), id(i, c + 1), F[i].f(c));
    link(id(i, R[i] + 1), T, inf);
  }
  for (int i = 1, u, v, d; i <= m; i++) {
    cin >> u >> v >> d;
    for (int c = L[v]; c <= R[v] + 1; c++)
      if (c + d >= L[u] && c + d <= R[u] + 1) link(id(u, c + d), id(v, c), inf);
  }
  Ans = n * (1ll << 25);
  while (BFS()) {
    for (int i = 1; i <= nod; i++) cur[i] = head[i];
    while (long long tmp = DFS(S, inf)) Ans -= tmp;
  }
  cout << Ans << endl;
}