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 inf = 1000000000;
struct MaxFlow {
  static const int MAX = 100005;
  int n, s, t;
  struct Edge {
    int to, cap, rev;
    Edge() {}
    Edge(int to, int cap) : to(to), cap(cap) {}
  };
  int bio[MAX];
  int cookie;
  int p[MAX];
  vector<Edge> G[MAX];
  MaxFlow(int n, int s, int t) : n(n), s(s), t(t) {
    memset(bio, 0, sizeof bio);
    cookie = 1;
  }
  void add_edge(int a, int b, int cap) {
    G[a].push_back(Edge(b, cap));
    G[b].push_back(Edge(a, 0));
    G[a].back().rev = G[b].size() - 1;
    G[b].back().rev = G[a].size() - 1;
  }
  int augment(int now, int flow) {
    if (now == t) return flow;
    if (bio[now] == cookie) return 0;
    bio[now] = cookie;
    int sent = 0;
    for (auto &e : G[now]) {
      if (e.cap == 0) continue;
      int send = augment(e.to, min(e.cap, flow));
      flow -= send;
      sent += send;
      e.cap -= send;
      G[e.to][e.rev].cap += send;
    }
    return sent;
  }
  int maxflow() {
    int ret = 0;
    int f;
    ++cookie;
    while ((f = augment(s, inf))) {
      ret += f;
      ++cookie;
    }
    return ret;
  }
};
const int MAXN = 55;
const int MAXM = 105;
int n, m;
int a[MAXN];
int b[MAXN];
int c[MAXN];
int l[MAXM];
int r[MAXM];
const int s = 0;
const int t = 1;
int id_alloc = 1;
map<pair<int, int>, int> id;
int get_id(int i, int x) {
  if (id.count({i, x})) return id[make_pair(i, x)];
  id[make_pair(i, x)] = ++id_alloc;
  return id_alloc;
}
int f(int i, int x) { return a[i] * x * x + b[i] * x + c[i]; }
int main(void) {
  scanf("%d%d", &n, &m);
  for (int i = (0); i < (n); ++i) scanf("%d%d%d", a + i, b + i, c + i);
  for (int i = (0); i < (n); ++i) scanf("%d%d", l + i, r + i);
  MaxFlow MF(n * 202, s, t);
  int ZERO = 0;
  for (int i = (0); i < (n); ++i)
    for (int x = l[i]; x <= r[i]; ++x) ZERO = min(ZERO, -f(i, x));
  ZERO = -ZERO;
  for (int i = (0); i < (n); ++i) {
    MF.add_edge(s, get_id(i, l[i] - 1), inf);
    MF.add_edge(get_id(i, r[i]), t, inf);
    for (int x = l[i]; x <= r[i]; ++x)
      MF.add_edge(get_id(i, x - 1), get_id(i, x), ZERO - f(i, x));
  }
  for (int i = (0); i < (m); ++i) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    --u;
    --v;
    for (int x = l[u] - 1; x <= r[u]; ++x)
      MF.add_edge(get_id(u, x), get_id(v, x - d), inf);
  }
  printf("%d\n", n * ZERO - MF.maxflow());
  return 0;
}