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;
template <class T>
void read(T &x) {
  x = 0;
  bool flag = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar()) flag |= ch == '-';
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - 48;
  flag ? x = ~x + 1 : 0;
}
int head[10010], cur[10010], tot = 1, q[10010], dep[10010];
struct edge {
  int v, c, next;
} e[200010];
inline void insert(int u, int v, int c) {
  e[++tot].v = v, e[tot].c = c, e[tot].next = head[u];
  head[u] = tot;
}
inline void add(int u, int v, int c) { insert(u, v, c), insert(v, u, 0); }
inline bool bfs(int S, int T) {
  memset(dep, 0, sizeof dep);
  dep[S] = 1;
  int l = 1, r = 1;
  q[1] = S;
  while (l <= r) {
    int u = q[l++];
    for (int i = head[u], v; i; i = e[i].next)
      if (e[i].c && !dep[v = e[i].v]) {
        dep[v] = dep[u] + 1, q[++r] = v;
        if (v == T) return 1;
      }
  }
  return 0;
}
int dfs(int u, int dist, int T) {
  if (u == T) return dist;
  int ret = 0;
  for (int &i = head[u], v; i; i = e[i].next)
    if (dep[v = e[i].v] == dep[u] + 1 && e[i].c) {
      int d = dfs(v, min(dist - ret, e[i].c), T);
      e[i].c -= d, e[i ^ 1].c += d, ret += d;
      if (ret == dist) return dist;
    }
  if (!ret) dep[u] = -1;
  return ret;
}
int dinic(int S, int T) {
  int ret = 0;
  memcpy(cur, head, sizeof head);
  while (bfs(S, T)) ret += dfs(S, 0x3f3f3f3f, T), memcpy(head, cur, sizeof cur);
  return ret;
}
int a[10010], b[10010], c[10010], l[10010], r[10010], node[10010];
inline int f(int x, int y) { return a[x] * y * y + b[x] * y + c[x]; }
inline int id(int x, int y) { return node[x] + y - l[x]; }
int main() {
  int n, m;
  read(n), read(m);
  for (int i = (1); i <= (n); i++) read(a[i]), read(b[i]), read(c[i]);
  int cnt = 0, M = 0;
  for (int i = (1); i <= (n); i++) {
    read(l[i]), read(r[i]);
    node[i] = cnt + 1, cnt += r[i] - l[i] + 2;
    for (int j = (l[i]); j <= (r[i]); j++) M = max(M, f(i, j));
  }
  int S = cnt + 1, T = S + 1;
  for (int i = (1); i <= (n); i++) {
    add(S, node[i], 0x3f3f3f3f);
    for (int j = (l[i]); j <= (r[i]); j++)
      add(id(i, j), id(i, j + 1), M - f(i, j));
    add(id(i, r[i] + 1), T, 0x3f3f3f3f);
  }
  while (m--) {
    int u, v, d;
    read(u), read(v), read(d);
    for (int i = (l[u]); i <= (r[u]); i++)
      if (i - d >= l[v] && i - d <= r[v] + 1)
        add(id(u, i), id(v, i - d), 0x3f3f3f3f);
  }
  printf("%d", M * n - dinic(S, T));
  return 0;
}