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>
enum { MAX = 50 * 3 * 100000, INF = 1000000000 };
struct Edge {
  int b, c, inv, next;
};
struct Flow {
  std::vector<int> G;
  std::vector<Edge> E;
  int alloc(int n) {
    int s = G.size();
    G.resize(s + n, -1);
    return s;
  }
  void push(int a, int b, int c) {
    int i = E.size(), j = i + 1;
    E.push_back(Edge{b, c, j, G[a]});
    E.push_back(Edge{a, 0, i, G[b]});
    G[a] = i;
    G[b] = j;
  }
  std::vector<int> L;
  std::vector<int> N;
  void bfs(int s) {
    L.assign(G.size(), -1);
    L[s] = 0;
    std::vector<int> Q{s};
    int qf = 0;
    while (qf < Q.size()) {
      int v = Q[qf++];
      for (int e = G[v]; e != -1; e = E[e].next)
        if (E[e].c && L[E[e].b] == -1) {
          L[E[e].b] = L[v] + 1;
          Q.push_back(E[e].b);
        }
    }
  }
  int dfs(int v, int t, int c) {
    if (v == t) return c;
    for (int &e = N[v]; e != -1; e = E[e].next)
      if (L[E[e].b] == L[v] + 1 && E[e].c > 0) {
        int c2 = dfs(E[e].b, t, std::min(c, E[e].c));
        if (c2) {
          E[e].c -= c2;
          E[E[e].inv].c += c2;
          return c2;
        }
      }
    return 0;
  }
  int go(int s, int t) {
    int res = 0;
    while (1) {
      bfs(s);
      if (L[t] == -1) return res;
      N.assign(G.begin(), G.end());
      while (1) {
        int c = dfs(s, t, INF);
        if (!c) break;
        res += c;
      }
    }
  }
  void print() {
    for (int v = 0; v < (G.size()); ++v) {
      printf("%d -> ", v);
      for (int e = G[v]; e != -1; e = E[e].next)
        printf("%d(c=%d) ", E[e].b, E[e].c);
      puts("");
    }
  }
};
struct Fun {
  int a, b, c, l, r;
  int operator()(int x) { return a * x * x + b * x + c; }
  int base;
};
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  std::vector<Fun> F(n);
  for (Fun &f : F) scanf("%d%d%d", &f.a, &f.b, &f.c);
  for (Fun &f : F) scanf("%d%d", &f.l, &f.r);
  Flow flow;
  int s = flow.alloc(1), t = flow.alloc(1);
  for (Fun &f : F) {
    f.base = flow.alloc(f.r - f.l + 2) - f.l;
    for (int x = (f.l); x <= (f.r); ++x)
      flow.push(f.base + x, f.base + x + 1, MAX - f(x));
    flow.push(s, f.base + f.l, INF);
    flow.push(f.base + f.r + 1, t, INF);
  }
  while (m--) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    u--;
    v--;
    for (int x = (F[u].l); x <= (F[u].r + 1); ++x)
      if (F[v].l <= x - d && x - d <= F[v].r + 1)
        flow.push(F[u].base + x, F[v].base + x - d, INF);
  }
  printf("%d\n", MAX * n - flow.go(s, t));
  return 0;
}