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;
inline int gi() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - 48;
    ch = getchar();
  }
  return x * f;
}
template <typename T>
inline bool Max(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool Min(T &a, T b) {
  return a > b ? a = b, 1 : 0;
}
const int N = 1e4 + 104, inf = 0x7fffffff, lim = 1e6;
int n, m;
int id[203][203], l[N], r[N];
struct Function {
  int a, b, c;
  int f(int x) {
    x -= 100;
    return a * x * x + b * x + c;
  }
} f[N];
namespace MF {
int edc = 1, ndc, st, ed;
int head[N], cur[N], dep[N];
struct edge {
  int lst, to, cap, flow;
} e[N * 20];
void Add(int a, int b, int c) {
  e[++edc] = (edge){head[a], b, c, 0}, head[a] = edc;
  e[++edc] = (edge){head[b], a, 0, 0}, head[b] = edc;
}
bool bfs() {
  queue<int> Q;
  for (int i = 1; i <= ndc; ++i) dep[i] = 0, cur[i] = head[i];
  dep[st] = 1;
  Q.push(st);
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int i = head[u], v = e[i].to; i; i = e[i].lst, v = e[i].to)
      if (e[i].cap - e[i].flow > 0 && !dep[v]) {
        dep[v] = dep[u] + 1;
        Q.push(v);
      }
  }
  return dep[ed];
}
int dfs(int u, int b) {
  if (u == ed || !b) return b;
  int flow = 0, f;
  for (int &i = cur[u], v = e[i].to; i; i = e[i].lst, v = e[i].to)
    if (e[i].cap - e[i].flow > 0 && dep[u] + 1 == dep[v]) {
      f = dfs(v, min(b, e[i].cap - e[i].flow));
      flow += f, e[i].flow += f, e[i ^ 1].flow -= f, b -= f;
      if (!b) break;
    }
  return flow;
}
int Maxflow() {
  int res = 0;
  while (bfs()) res += dfs(st, inf);
  return res;
}
}  // namespace MF
int main() {
  using namespace MF;
  n = gi(), m = gi();
  for (int i = 1; i <= n; ++i) f[i].a = gi(), f[i].b = gi(), f[i].c = gi();
  for (int i = 1; i <= n; ++i) {
    l[i] = gi() + 100, r[i] = gi() + 100;
    for (int j = l[i]; j <= r[i] + 1; ++j) id[i][j] = ++ndc;
    for (int j = l[i]; j <= r[i]; ++j)
      Add(id[i][j], id[i][j + 1], lim - f[i].f(j));
  }
  st = ++ndc, ed = ++ndc;
  for (int i = 1; i <= n; ++i)
    Add(st, id[i][l[i]], inf), Add(id[i][r[i] + 1], ed, inf);
  while (m--) {
    int u = gi(), v = gi(), d = gi();
    for (int i = l[v]; i <= r[v]; ++i) {
      int p = i + d + 1;
      if (l[u] <= p && p <= r[u])
        Add(id[u][p], id[v][i + 1], inf);
      else if (p < l[u])
        Add(id[u][l[u]], id[v][i + 1], inf);
    }
  }
  printf("%d\n", n * lim - Maxflow());
  return 0;
}