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;
struct DinicFlow {
  struct Edge {
    Edge(int _to, int _cap, int _next) : to(_to), cap(_cap), next(_next){};
    int to, cap, next;
  };
  int n, s, t, fl;
  vector<int> first, d, q, ptr;
  vector<bool> used;
  vector<Edge> a;
  DinicFlow(int _n, int _s, int _t) : n(_n), s(_s), t(_t) {
    first.resize(n, -1);
  };
  void addEdge(int u, int v, int cap, bool oriented = true) {
    a.push_back(Edge(v, cap, first[u]));
    first[u] = a.size() - 1;
    a.push_back(Edge(u, oriented ? 0 : cap, first[v]));
    first[v] = a.size() - 1;
  }
  bool bfs() {
    d.clear();
    d.resize(n, -1);
    d[s] = 0;
    q.clear();
    q.push_back(s);
    for (int i = 0; i < q.size(); ++i) {
      int v = q[i];
      for (int e = first[v]; e != -1; e = a[e].next) {
        int u = a[e].to;
        if (d[u] == -1 && a[e].cap) {
          d[u] = d[v] + 1;
          q.push_back(u);
        }
      }
    }
    return d[t] != -1;
  }
  int dfs(int v, int push = 1000000000) {
    if (v == t || push <= 0) {
      fl += push;
      return push;
    }
    int pushed = 0;
    for (int e = ptr[v]; e != -1; e = a[e].next) {
      int u = a[e].to;
      ptr[v] = a[e].next;
      if (d[v] + 1 == d[u]) {
        if (int val = dfs(u, min(push - pushed, a[e].cap))) {
          pushed += val;
          a[e].cap -= val;
          a[e ^ 1].cap += val;
        }
      }
    }
    return pushed;
  }
  int flow() {
    fl = 0;
    while (bfs()) {
      ptr = first;
      dfs(s);
    }
    return fl;
  }
};
const int MAX = 10000000;
int a[51], b[51], c[51], l[51], r[51], L[51], R[51];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
  }
  int s = 0, t = 201 * n + 1, last = 0;
  DinicFlow flow(t + 1, s, t);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &l[i], &r[i]);
    L[i] = l[i];
    R[i] = r[i];
    flow.addEdge(s, last + 1, 10 * MAX);
    for (int j = l[i]; j <= r[i]; ++j) {
      flow.addEdge(last + j - l[i] + 1, last + j - l[i] + 2,
                   MAX - (a[i] * j * j + b[i] * j + c[i]));
    }
    flow.addEdge(last + r[i] - l[i] + 2, t, 10 * MAX);
    r[i] = last + r[i] - l[i] + 2;
    l[i] = last + 2;
    last = r[i];
  }
  for (int i = 0; i < m; ++i) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    --u;
    --v;
    for (int j = L[u] - 1; j <= R[u]; ++j) {
      int ind = (j - d - L[v]) + l[v];
      if (l[v] - 1 <= ind && ind <= r[v]) {
        flow.addEdge(l[u] + (j - L[u]), ind, 10 * MAX);
      }
    }
  }
  printf("%d\n", MAX * n - flow.flow());
  return 0;
}