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;
void getint(int &v) {
  char ch, fu = 0;
  for (ch = '*'; (ch < '0' || ch > '9') && ch != '-'; ch = getchar())
    ;
  if (ch == '-') fu = 1, ch = getchar();
  for (v = 0; ch >= '0' && ch <= '9'; ch = getchar()) v = v * 10 + ch - '0';
  if (fu) v = -v;
}
const int INF = 1e9 + 10;
int q[500010], a[500010], p[500010], now, ans, M[500010], b[500010], c[500010],
    hed[500010], d[500010], S, T, n, m, l[500010], r[500010], cap[500010],
    too[500010], nxt[500010], nedge;
void ae(int x, int y, int f) {
  nxt[++nedge] = hed[x];
  hed[x] = nedge;
  too[nedge] = y;
  cap[nedge] = f;
}
void add(int x, int y, int f) { ae(x, y, f), ae(y, x, 0); }
bool build() {
  int he = 0, ta = 1;
  q[1] = S;
  for (int i = S; i <= T; i++) d[i] = -1;
  d[S] = 0;
  while (he != ta) {
    int x = q[++he];
    for (int i = hed[x]; i; i = nxt[i]) {
      int y = too[i];
      if (d[y] != -1 || !cap[i]) continue;
      d[y] = d[x] + 1;
      if (y == T) return 1;
      q[++ta] = y;
    }
  }
  return 0;
}
int fnd(int x, int flo) {
  if (x == T) return flo;
  int ret, w = 0;
  for (int i = hed[x]; i && w < flo; i = nxt[i]) {
    int y = too[i];
    if (cap[i] && d[y] == d[x] + 1 && (ret = fnd(y, min(flo - w, cap[i]))))
      cap[i] -= ret, cap[i ^ 1] += ret, w += ret;
  }
  d[x] = -1;
  return w;
}
int dinic() {
  int ans = 0;
  while (build())
    while (1) {
      int flo = fnd(S, INF);
      if (!flo) break;
      ans += flo;
    }
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i] >> c[i];
  }
  for (int i = 1; i <= n; i++) {
    getint(l[i]), getint(r[i]);
    M[i] = -INF;
    for (int x = l[i]; x <= r[i]; x++)
      M[i] = max(M[i], a[i] * x * x + b[i] * x + c[i]);
    T += (r[i] - l[i] + 1) * 2;
  }
  S = 0;
  ++T;
  nedge = 1;
  now = 0;
  for (int i = 1; i <= n; i++) {
    ans += M[i];
    p[i] = now;
    add(S, now + 1, INF);
    for (int x = l[i]; x <= r[i]; x++) {
      int t = M[i] - (a[i] * x * x + b[i] * x + c[i]);
      if (x > l[i]) add(now, now + 1, INF);
      ++now;
      add(now, now + 1, t);
      ++now;
    }
    add(now, T, INF);
  }
  for (int i = 1; i <= m; i++) {
    int u, v, d;
    cin >> u >> v >> d;
    now = p[u] - 1;
    for (int x = l[u]; x <= r[u]; x++) {
      now += 2;
      int t = x - d;
      if (t < l[v]) continue;
      if (t > r[v])
        add(now, T, INF);
      else
        add(now, p[v] + 1 + (t - l[v]) * 2, INF);
    }
  }
  ans -= dinic();
  cout << ans << endl;
  return 0;
}