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 S, T;
struct dat {
  int to, rev, f, cap;
  dat(int to = 0, int rev = 0, int f = 0, int cap = 0)
      : to(to), rev(rev), f(f), cap(cap) {}
};
vector<dat> v[11000];
int a[11000], b[11000], c[11000], l[11000], r[11000], st[11000];
int dist[11000];
queue<int> q;
int work[11000];
void add(int x, int y, int cap) {
  int n = v[x].size();
  int m = v[y].size();
  v[x].push_back(dat(y, m, 0, cap));
  v[y].push_back(dat(x, n, 0, 0));
}
bool bfs() {
  for (int i = 0; i <= T; i++) dist[i] = 1000000000;
  while (!q.empty()) q.pop();
  q.push(S);
  dist[S] = 0;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = 0; i < v[x].size(); i++) {
      dat aa = v[x][i];
      if (aa.f == aa.cap) continue;
      int y = aa.to;
      if (dist[y] != 1000000000) continue;
      dist[y] = dist[x] + 1;
      q.push(y);
    }
  }
  return dist[T] != 1000000000;
}
int dfs(int x, int f) {
  if (x == T) return f;
  for (; work[x] < v[x].size(); work[x]++) {
    dat aa = v[x][work[x]];
    if (aa.f == aa.cap) continue;
    int y = aa.to;
    if (dist[y] != dist[x] + 1) continue;
    int d = dfs(y, min(f, aa.cap - aa.f));
    if (d) {
      v[x][work[x]].f += d;
      v[y][aa.rev].f -= d;
      return d;
    }
  }
  return 0;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
  }
  T = 0;
  S = 0;
  int val = 400000;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    ++T;
    st[i] = T;
    add(S, T, 1000000000);
    for (int j = l[i]; j <= r[i]; j++) {
      int xx = a[i] * j * j + b[i] * j + c[i];
      add(T, T + 1, val - xx);
      T++;
    }
  }
  ++T;
  for (int i = 1; i <= n; i++) {
    add(st[i] + r[i] - l[i] + 1, T, 1000000000);
  }
  for (int i = 1; i <= m; i++) {
    int x, y, d;
    scanf("%d%d%d", &x, &y, &d);
    for (int j = l[x]; j <= r[x]; j++) {
      int xx = j - d;
      if (xx >= l[y] && xx <= r[y] + 1) {
        add(st[x] + j - l[x], st[y] + xx - l[y], 1000000000);
      }
    }
  }
  int ans = val * n;
  while (bfs()) {
    for (int i = 0; i <= T; i++) work[i] = 0;
    while (1) {
      int d = dfs(S, 1000000000);
      if (!d) break;
      ans -= d;
    }
  }
  printf("%d\n", ans);
  return 0;
}