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;
int tot, t[2000010], p[2000010], nt[2000010];
long long ans, w[2000010];
void init(int n) {
  ans = 0;
  tot = 0;
  for (int i = (1); i <= (n); ++i) t[i] = -1;
}
void lk(int x, int y, long long z) {
  nt[tot] = t[x];
  p[tot] = y;
  w[tot] = z;
  t[x] = tot++;
  nt[tot] = t[y];
  p[tot] = x;
  w[tot] = 0;
  t[y] = tot++;
}
int dl[2000010], h[2000010], cur[2000010];
long long dfs(int n, int x, long long y) {
  if (x == n) return y;
  long long t0 = y;
  for (; ~cur[x]; cur[x] = nt[cur[x]])
    if (w[cur[x]] && h[p[cur[x]]] > h[x]) {
      long long z = dfs(n, p[cur[x]], min(w[cur[x]], y));
      w[cur[x]] -= z;
      w[cur[x] ^ 1] += z;
      y -= z;
      if (!y) break;
    }
  return t0 - y;
}
void dinic(int n) {
  for (;;) {
    for (int i = (1); i <= (n); ++i) h[i] = 0, cur[i] = t[i];
    int s = 1, ta = 1;
    dl[1] = h[1] = 1;
    for (; s <= ta; s++)
      for (int i = t[dl[s]]; ~i; i = nt[i])
        if (w[i] && !h[p[i]]) {
          h[p[i]] = h[dl[s]] + 1;
          dl[++ta] = p[i];
        }
    if (!h[n]) return;
    ans += dfs(n, 1, 2000000000000000000ll);
  }
}
int n, m, A[2000010], B[2000010], C[2000010];
int main() {
  scanf("%d%d", &n, &m);
  init(n * 201 + 2);
  for (int i = (1); i <= (n); ++i) scanf("%d%d%d", &A[i], &B[i], &C[i]);
  for (int i = (1); i <= (n); ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    for (int j = (-100); j <= (100); ++j)
      lk(j == -100 ? 1 : (i - 1) * 201 + j + 101, (i - 1) * 201 + j + 102,
         j < l || j > r ? 2000000000000000000ll
                        : 200000 - (A[i] * j * j + B[i] * j + C[i]));
    lk(i * 201 + 1, n * 201 + 2, 2000000000000000000ll);
  }
  for (int i = (1); i <= (m); ++i) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    for (int j = (-300); j <= (100); ++j)
      if (j - z >= -100) {
        if (j >= -100) {
          if (j - z > 100)
            lk((x - 1) * 201 + j + 102, n * 201 + 2, 2000000000000000000ll);
          else
            lk((x - 1) * 201 + j + 102, (y - 1) * 201 + j - z + 102,
               2000000000000000000ll);
        } else if (j - z <= 100)
          lk(1, (y - 1) * 201 + j - z + 102, 2000000000000000000ll);
      }
  }
  dinic(n * 201 + 2);
  printf("%lld\n", 200000 * n - ans);
}