434_D. Nanami's Power Plant

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.

Example 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

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


Example output:
46

9


Tags: flows
Difficulty rating: 2900

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 10005, M = N * 3, Inf = 0x3f3f3f3f;
int n, m, s, t;
int a[N], b[N], c[N], l[N], r[N], idx, st[N];
int eid = 1, he[N], nw[N], d[N], q[N];
struct edge {
  int v, nx, c;
} e[M * 2];
void add(int u, int v, int c) {
  e[++eid] = {v, he[u], c}, he[u] = eid;
  e[++eid] = {u, he[v], 0}, he[v] = eid;
}
int bfs() {
  memset(d, 0, sizeof d), memcpy(nw, he, sizeof he);
  q[1] = s, d[s] = 1;
  for (int l = 1, r = 1; l <= r; l++) {
    int u = q[l];
    for (int i = he[u], v; v = e[i].v; i = e[i].nx)
      if (e[i].c && !d[v]) q[++r] = v, d[v] = d[u] + 1;
  }
  return d[t];
}
long long dfs(int u, long long lim, long long re = 0) {
  if (u == t) return lim;
  for (int &i = nw[u], v; v = e[i].v; i = e[i].nx)
    if (e[i].c && d[v] == d[u] + 1) {
      long long t = dfs(v, min((long long)e[i].c, lim));
      e[i].c -= t, e[i ^ 1].c += t, re += t, lim -= t;
      if (!lim) break;
    }
  if (lim) d[u] = 0;
  return re;
}
int main() {
  cin >> n >> m;
  s = ++idx, t = ++idx;
  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]);
    add(s, ++idx, Inf * 2), st[i] = idx + 1 - l[i];
    for (int j = (l[i]); j <= (r[i]); j++)
      add(idx, idx + 1, Inf - a[i] * j * j - b[i] * j - c[i]), idx++;
    add(idx, t, Inf * 2);
  }
  int u, v, d;
  while (m--) {
    scanf("%d%d%d", &u, &v, &d);
    for (int i = (max(l[u] - d, l[v]) - 1); i <= (min(r[u] - d, r[v])); i++)
      add(st[u] + i + d, st[v] + i, Inf * 2);
  }
  long long as = (long long)Inf * n;
  while (bfs()) as -= dfs(s, Inf * 2);
  cout << as;
  return 0;
}