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 = 15000;
int level[maxn], que[maxn];
int head[maxn], lon;
int aa[70], bb[70], cc[70];
int l[70], r[70];
int start[70];
struct edge {
  int next, to, c;
} e[500000];
void edgeini() {
  memset(head, -1, sizeof(head));
  lon = -1;
}
void edgemake(int from, int to, int c) {
  e[++lon].c = c;
  e[lon].to = to;
  e[lon].next = head[from];
  head[from] = lon;
}
void make(int from, int to, int c) {
  edgemake(from, to, c);
  edgemake(to, from, 0);
}
bool makelevel(int s, int t) {
  memset(level, 0, sizeof(level));
  int front = 1, end = 0;
  que[++end] = s;
  level[s] = 1;
  while (front <= end) {
    int u = que[front++];
    if (u == t) return true;
    for (int k = head[u]; k != -1; k = e[k].next) {
      int v = e[k].to;
      if (!level[v] && e[k].c) {
        que[++end] = v;
        level[v] = level[u] + 1;
      }
    }
  }
  return false;
}
int dfs(int now, int t, int maxf) {
  if (now == t || maxf == 0) return maxf;
  int ret = 0;
  for (int k = head[now]; k != -1; k = e[k].next) {
    int u = e[k].to;
    if (level[u] == level[now] + 1 && e[k].c) {
      int f = dfs(u, t, min(e[k].c, maxf - ret));
      e[k].c -= f;
      e[k ^ 1].c += f;
      ret += f;
      if (ret == maxf) return ret;
    }
  }
  if (ret == 0) level[now] = 0;
  return ret;
}
int maxflow(int s, int t) {
  int ret = 0;
  while (makelevel(s, t)) {
    ret += dfs(s, t, 1 << 30);
  }
  return ret;
}
int fun(int p, int x) { return aa[p] * x * x + bb[p] * x + cc[p]; }
int id(int p, int x) { return start[p] + x - l[p]; }
int main() {
  int n, m;
  int i, j;
  scanf("%d%d", &n, &m);
  edgeini();
  int MAX = -1 << 30;
  int cnt = 0;
  for (i = 1; i <= n; i++) scanf("%d%d%d", &aa[i], &bb[i], &cc[i]);
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    start[i] = cnt + 1;
    cnt = cnt + r[i] - l[i] + 2;
    for (j = l[i]; j <= r[i]; j++) MAX = max(MAX, fun(i, j));
  }
  for (i = 1; i <= n; i++) {
    make(0, start[i], 1 << 30);
    for (j = l[i]; j <= r[i]; j++)
      make(id(i, j), id(i, j + 1), MAX - fun(i, j));
    make(id(i, r[i] + 1), cnt + 1, 1 << 30);
  }
  int u, v, d;
  while (m--) {
    scanf("%d%d%d", &u, &v, &d);
    for (i = l[u]; i <= r[u]; i++)
      if (i - d >= l[v] && i - d <= r[v] + 1)
        make(id(u, i), id(v, i - d), 1 << 30);
  }
  int ans = maxflow(0, cnt + 1);
  printf("%d\n", MAX * n - ans);
}