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 maxn = 50 + 14, maxf = 2e5;
namespace F {
const int maxn = 2e4, maxm = maxn * 2, inf = 1e9, so = maxn - 1, si = maxn - 2;
int ptr[maxn], head[maxn], prv[maxm], to[maxm], cap[maxm], d[maxn], q[maxn], ec;
void init() {
  memset(head, -1, sizeof head);
  ec = 0;
}
void add(int v, int u, int vu, int uv = 0) {
  to[ec] = u, prv[ec] = head[v], cap[ec] = vu, head[v] = ec++;
  to[ec] = v, prv[ec] = head[u], cap[ec] = uv, head[u] = ec++;
}
bool bfs() {
  memset(d, 63, sizeof d);
  d[so] = 0;
  int h = 0, t = 0;
  q[t++] = so;
  while (h < t) {
    int v = q[h++];
    for (int e = head[v]; e >= 0; e = prv[e])
      if (cap[e] && d[to[e]] > d[v] + 1) {
        d[to[e]] = d[v] + 1;
        q[t++] = to[e];
        if (to[e] == si) return 1;
      }
  }
  return 0;
}
int dfs(int v, int f = inf) {
  if (v == si || f == 0) return f;
  int r = 0;
  for (int &e = ptr[v]; e >= 0; e = prv[e])
    if (d[v] == d[to[e]] - 1) {
      int x = dfs(to[e], min(f, cap[e]));
      f -= x, r += x;
      cap[e] -= x, cap[e ^ 1] += x;
      if (!f) break;
    }
  return r;
}
int mf() {
  int ans = 0;
  while (bfs()) {
    memcpy(ptr, head, sizeof ptr);
    ans += dfs(so);
  }
  return ans;
}
};  // namespace F
int n, m, x, a[maxn], b[maxn], c[maxn];
int id(int i, int x) {
  assert(x >= -101 && x <= 100);
  return i * 202 + x + 101;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  F::init();
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i] >> b[i] >> c[i];
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    F::add(F::so, id(i, l - 1), F::inf);
    for (int j = l; j <= r; j++)
      F::add(id(i, j - 1), id(i, j), maxf - (a[i] * j * j + b[i] * j + c[i]));
    F::add(id(i, r), F::si, F::inf);
  }
  for (int i = 0; i < m; i++) {
    int u, v, d;
    cin >> u >> v >> d;
    u--, v--;
    for (int j = max(-101, -101 + d); j <= min(100, 100 + d); j++)
      F::add(id(u, j), id(v, j - d), F::inf);
  }
  cout << n * maxf - F::mf() << '\n';
}