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 * 201 + 10;
const int maxm = (60 * 210 + 100) * 2;
const int INF = 0x3f3f3f3f;
int n, m, en, q[maxn], depth[maxn], l[60], r[60], start[60], A[60], B[60],
    C[60], s, t;
struct edge {
  int e, f;
  edge *next, *op;
} * v[maxn], ed[maxm];
void add_edge(int s, int e, int f) {
  en++;
  ed[en].next = v[s];
  v[s] = ed + en;
  v[s]->e = e;
  v[s]->f = f;
  en++;
  ed[en].next = v[e];
  v[e] = ed + en;
  v[e]->e = s;
  v[e]->f = 0;
  v[s]->op = v[e];
  v[e]->op = v[s];
}
bool bfs() {
  int front = 1, tail = 1;
  q[1] = s;
  memset(depth, 0, sizeof(depth));
  depth[s] = 1;
  for (; front <= tail;) {
    int now = q[front++];
    for (edge *e = v[now]; e; e = e->next)
      if (e->f && !depth[e->e]) {
        depth[e->e] = depth[now] + 1;
        if (e->e == t) return true;
        q[++tail] = e->e;
      }
  }
  return false;
}
int dfs(int now, int cur_flow) {
  if (now == t) return cur_flow;
  int rest = cur_flow;
  for (edge *e = v[now]; e && rest; e = e->next)
    if (e->f && depth[e->e] == depth[now] + 1) {
      int new_flow = dfs(e->e, min(e->f, rest));
      e->f -= new_flow;
      e->op->f += new_flow;
      rest -= new_flow;
    }
  if (rest == cur_flow) depth[now] = -1;
  return cur_flow - rest;
}
int dinic() {
  int ans = 0;
  while (bfs()) ans += dfs(s, INF);
  return ans;
}
int f(int x, int y) { return y * y * A[x] + y * B[x] + C[x]; }
int main() {
  scanf("%d%d", &n, &m);
  for (int a = 1; a <= n; a++) scanf("%d%d%d", &A[a], &B[a], &C[a]);
  int maxv = -INF;
  int tot = 0;
  for (int a = 1; a <= n; a++) {
    scanf("%d%d", &l[a], &r[a]);
    start[a] = tot + 1;
    tot += r[a] - l[a] + 2;
    for (int b = l[a]; b <= r[a]; b++) maxv = max(maxv, f(a, b));
  }
  maxv++;
  s = 0;
  t = tot + 1;
  for (int a = 1; a <= n; a++) {
    add_edge(s, start[a], INF);
    for (int b = l[a]; b <= r[a]; b++)
      add_edge((b - l[a] + start[a]), (b + 1 - l[a] + start[a]),
               maxv - f(a, b));
    add_edge((r[a] + 1 - l[a] + start[a]), t, INF);
  }
  for (int a = 1; a <= m; a++) {
    int s, e, d;
    scanf("%d%d%d", &s, &e, &d);
    for (int b = l[s]; b <= r[s]; b++)
      if (b - d >= l[e] && b - d <= r[e] + 1)
        add_edge((b - l[s] + start[s]), (b - d - l[e] + start[e]), INF);
  }
  printf("%d\n", maxv * n - dinic());
  return 0;
}