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;
long long n, m, S, T, a[101], b[101], c[101], tot = 1, e[200001], nt[200001],
                                              hd[200001], val[200001], d[11001],
                                              q[200001];
void build(long long x, long long y, long long w) {
  tot++;
  e[tot] = y;
  val[tot] = w;
  nt[tot] = hd[x];
  hd[x] = tot;
  tot++;
  e[tot] = x;
  val[tot] = 0;
  nt[tot] = hd[y];
  hd[y] = tot;
}
long long get(long long x, long long y) {
  return -(a[x] * y * y + b[x] * y + c[x]) + 1e8;
}
long long bfs() {
  long long i, u, l = 0, r = 1;
  memset(d, 0, sizeof(d));
  d[S] = 1;
  q[1] = S;
  while (l < r) {
    u = q[++l];
    for (i = hd[u]; i; i = nt[i])
      if (!d[e[i]] && val[i]) {
        d[e[i]] = d[u] + 1;
        q[++r] = e[i];
      }
  }
  return d[T] > 0;
}
long long dfs(long long x, long long flow) {
  long long i, sum = 0, hh;
  if (x == T) return flow;
  for (i = hd[x]; i; i = nt[i])
    if (val[i] && d[e[i]] == d[x] + 1) {
      hh = dfs(e[i], min(flow, val[i]));
      flow -= hh;
      sum += hh;
      val[i] -= hh;
      val[i ^ 1] += hh;
      if (!flow) return sum;
    }
  if (!sum) d[x] = 0;
  return sum;
}
long long dinic() {
  long long flow = 0;
  while (bfs()) {
    flow += dfs(S, 1e16);
  }
  return flow;
}
int main() {
  long long i, j, x, y, w;
  scanf("%lld%lld", &n, &m);
  S = n * 202 + 1;
  T = S + 1;
  for (i = 1; i <= n; i++) {
    scanf("%lld%lld%lld", &a[i], &b[i], &c[i]);
  }
  for (i = 1; i <= n; i++) {
    scanf("%lld%lld", &x, &y);
    x += 101;
    y += 101;
    build(S, (i - 1) * 202 + 1, 1e16);
    for (j = 1; j < x; j++)
      build((i - 1) * 202 + j, (i - 1) * 202 + j + 1, 1e16);
    for (j = x; j <= y; j++)
      build((i - 1) * 202 + j, (i - 1) * 202 + j + 1, get(i, j - 101));
    for (j = y + 1; j < 202; j++)
      build((i - 1) * 202 + j, (i - 1) * 202 + j + 1, 1e16);
    build((i - 1) * 202 + 202, T, 1e16);
  }
  for (i = 1; i <= m; i++) {
    scanf("%lld%lld%lld", &x, &y, &w);
    for (j = 1; j + w <= 202; j++)
      build((x - 1) * 202 + j + w, (y - 1) * 202 + j, 1e16);
  }
  printf("%lld\n", -(dinic() - (100000000) * n));
}