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 p1 {
  int a, b, c, l, r;
} p2[11005];
int a[60][300];
struct pi {
  int to;
  int cost;
  int rev;
} pp;
vector<pi> g[11005];
queue<int> q;
int s, t;
int line[11005], leve[11005];
int min(int a, int b) {
  int p;
  p = a;
  if (b < a) p = b;
  return p;
}
void add(int a, int b, int cos) {
  pp.to = b;
  pp.cost = cos;
  pp.rev = (int)g[b].size();
  g[a].push_back(pp);
  pp.to = a;
  pp.cost = 0;
  pp.rev = (int)g[a].size() - 1;
  g[b].push_back(pp);
  return;
}
void bfs(void) {
  q.push(s);
  int p, f, i;
  while (!q.empty()) {
    p = q.front();
    q.pop();
    f = (int)g[p].size();
    for (i = 0; i < f; i++) {
      pi &e = g[p][i];
      if (e.cost > 0 && line[e.to] < 0) {
        line[e.to] = line[p] + 1;
        q.push(e.to);
      }
    }
  }
  return;
}
int dfs(int v, int f) {
  int p;
  if (v == t) return f;
  for (int &i = leve[v]; i < g[v].size(); i++) {
    pi &e = g[v][i];
    if (line[v] < line[e.to] && e.cost > 0) {
      p = dfs(e.to, min(f, e.cost));
      if (p > 0) {
        e.cost -= p;
        g[e.to][e.rev].cost += p;
        return p;
      }
    }
  }
  return 0;
}
long long dinic() {
  int f;
  long long flow = 0;
  while (1) {
    memset(line, -1, sizeof(line));
    memset(leve, 0, sizeof(leve));
    line[s] = 0;
    bfs();
    if (line[t] < 0) return flow;
    f = dfs(s, 100000000);
    if (f == 0) continue;
    flow += f;
    while ((f = dfs(s, 100000000)) > 0) {
      flow += f;
    }
  }
}
int tot = 0;
int get(p1 p, int x) { return x * x * p.a + x * p.b + p.c; }
int main() {
  int i, j, n, m;
  cin >> n >> m;
  for (i = 1; i <= n; i++) scanf("%d%d%d", &p2[i].a, &p2[i].b, &p2[i].c);
  for (i = 1; i <= n; i++) scanf("%d%d", &p2[i].l, &p2[i].r);
  s = 0;
  t = 11000;
  for (i = 1; i <= n; i++) {
    a[i][0] = ++tot;
    add(0, tot, 100000000);
    for (j = 1; j <= p2[i].r - p2[i].l + 1; j++) {
      a[i][j] = ++tot;
      add(tot - 1, tot, 100000000 / 10 - get(p2[i], p2[i].l + j - 1));
    }
    add(tot, t, 100000000);
  }
  for (i = 0; i < m; i++) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    for (int j = p2[u].l - 1; j <= p2[u].r; j++) {
      if ((j - d) >= p2[v].l - 1 && (j - d) <= p2[v].r) {
        add(a[u][j - p2[u].l + 1], a[v][j - d - p2[v].l + 1], 100000000);
      }
    }
  }
  long long p = dinic();
  p = ((long long)n * (100000000 / 10)) - p;
  cout << p << endl;
}