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;
struct point {
  int next, to, w;
} ar[65000];
int a1[201], a2[201], a3[201], dep[11000], l1[201], r1[201], id[201][201],
    T[11000], tot, art(1), head[11000], a4[201][201], ps, pt;
inline bool bfs();
inline int f(int, int), dfs(int, int);
inline void read(int&), link(int, int, int), Min(int&, int);
int main() {
  int n, m, ans(0);
  read(n), read(m);
  for (int i(1); i <= n; ++i) read(a1[i]), read(a2[i]), read(a3[i]);
  for (int i(1); i <= n; ++i) read(l1[i]), read(r1[i]);
  memset(a4, 0x3f, sizeof(a4));
  for (int i(1), u, v, d; i <= m; ++i)
    read(u), read(v), read(d), Min(a4[u][v], d);
  for (int i(1), j; i <= n; ++i)
    for (j = l1[i]; j <= r1[i]; ++j) id[i][j + 100] = ++tot;
  ps = ++tot, pt = ++tot;
  for (int i(1), j, k; i <= n; ++i) {
    link(ps, id[i][l1[i] + 100], 0x3f3f3f3f);
    for (j = l1[i]; j <= r1[i]; ++j) {
      link(id[i][j + 100], j ^ r1[i] ? id[i][j + 1 + 100] : pt,
           201000 - f(i, j));
      for (k = 1; k <= n; ++k) {
        if (k == i || j - a4[i][k] <= l1[k]) continue;
        if (j - a4[i][k] > r1[k])
          link(id[i][j + 100], pt, 0x3f3f3f3f);
        else
          link(id[i][j + 100], id[k][j - a4[i][k] + 100], 0x3f3f3f3f);
      }
    }
    ans += 201000;
  }
  int flow, maxflow(0);
  while (bfs())
    while (flow = dfs(ps, 0x3f3f3f3f)) maxflow += flow;
  return printf("%d", ans - maxflow), 0;
}
inline int dfs(int x, int flow) {
  if (x == pt) return flow;
  int r(flow);
  for (int i(head[x]), j; i && r; i = ar[i].next)
    if (ar[i].w && dep[ar[i].to] == dep[x] + 1) {
      j = dfs(ar[i].to, min(r, ar[i].w));
      if (!j) {
        dep[ar[i].to] = 0;
        continue;
      }
      ar[i].w -= j, ar[i ^ 1].w += j, r -= j;
    }
  return flow - r;
}
inline bool bfs() {
  memset(dep, 0, tot + 1 << 2);
  int L(0), R(1);
  dep[T[1] = ps] = 1;
  while (L++ < R)
    for (int i(head[T[L]]); i; i = ar[i].next)
      if (ar[i].w && !dep[ar[i].to]) {
        dep[ar[i].to] = dep[T[L]] + 1;
        if ((T[++R] = ar[i].to) == pt) return true;
      }
  return false;
}
inline int f(int i, int x) { return a1[i] * x * x + a2[i] * x + a3[i]; }
inline void Min(int& x, int y) { x = x < y ? x : y; }
inline void link(int u, int v, int w) {
  ar[++art] = {head[u], v, w}, head[u] = art;
  ar[++art] = {head[v], u, 0}, head[v] = art;
}
inline void read(int& x) {
  x ^= x;
  register char c;
  while (c = getchar(), c == ' ' || c == '\n' || c == '\r')
    ;
  register bool check(false);
  if (c == '-') check = true, c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (check) x = ~x + 1;
}