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 n, m, a[111], b[111], c[111], l[111], r[111], id[111][111 * 2], cnt,
    maxx = -0x3f3f3f3f, ed = 11100;
int first[11111], vis[11111], v[88888], w[88888], nxt[88888], tot, U, V, D, jy,
    ans;
int func(int x, int y) { return a[x] * y * y + b[x] * y + c[x]; }
void Add(int x, int y, int z) {
  w[tot] = z, v[tot] = y, nxt[tot] = first[x], first[x] = tot++;
}
void add(int x, int y, int z) { Add(x, y, z), Add(y, x, 0); }
bool tell() {
  memset(vis, -1, sizeof(vis)), vis[0] = 0;
  queue<int> q;
  q.push(0);
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    for (int i = first[t]; ~i; i = nxt[i])
      if (w[i] && vis[v[i]] == -1) vis[v[i]] = vis[t] + 1, q.push(v[i]);
  }
  return vis[ed] != -1;
}
int zeng(int x, int y) {
  if (x == ed) return y;
  int r = 0;
  for (int i = first[x]; ~i && y > r; i = nxt[i])
    if (w[i] && vis[v[i]] == vis[x] + 1) {
      int t = zeng(v[i], min(w[i], y - r));
      w[i] -= t, w[i ^ 1] += t, r += t;
    }
  if (!r) vis[x] = -1;
  return r;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d%d%d", &a[i], &b[i], &c[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    for (int j = l[i]; j <= r[i]; j++) maxx = max(maxx, func(i, j));
  }
  maxx++;
  memset(first, -1, sizeof(first));
  for (int i = 1; i <= n; i++) {
    for (int j = l[i]; j <= r[i] + 1; j++) id[i][j + 100] = ++cnt;
    for (int j = l[i]; j <= r[i]; j++)
      add(id[i][j + 100], id[i][j + 101], maxx - func(i, j));
    add(0, id[i][l[i] + 100], 0x3f3f3f3f),
        add(id[i][r[i] + 101], ed, 0x3f3f3f3f);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &U, &V, &D);
    for (int j = l[U]; j <= r[U] + 1; j++)
      if (j - D >= l[V] && j - D <= r[V] + 1)
        add(id[U][j + 100], id[V][j - D + 100], 0x3f3f3f3f);
  }
  while (tell())
    while (jy = zeng(0, 0x3f3f3f3f)) ans += jy;
  printf("%d\n", maxx * n - ans);
}