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 long long read() {
  long long w = 0, x = 0;
  char c = getchar();
  while (!isdigit(c)) w |= c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();
  return w ? -x : x;
}
namespace star {
const long long maxn = 10010, maxm = 1e6 + 10, S = 10009, T = 10008,
                INF = 0x3f3f3f3f3f3f3f3f, lim = 1000000000000;
long long n, m, l[maxn], r[maxn];
struct function {
  long long a, b, c;
  inline long long operator()(long long x) const {
    return a * x * x + b * x + c;
  }
} f[maxn];
long long ecnt = 1, head[maxn], cur[maxn], to[maxm << 1], nxt[maxm << 1],
          w[maxm << 1], dis[maxn];
inline void addedge(long long a, long long b, long long c) {
  to[++ecnt] = b, nxt[ecnt] = head[a], head[a] = ecnt, w[ecnt] = c;
  to[++ecnt] = a, nxt[ecnt] = head[b], head[b] = ecnt, w[ecnt] = 0;
}
inline bool bfs() {
  memset(dis, -1, sizeof dis);
  static long long q[maxn];
  long long hd = 0, tl;
  dis[S] = 0;
  q[tl = 1] = S;
  while (hd < tl) {
    long long x = q[++hd];
    for (long long u, i = head[x]; i; i = nxt[i])
      if (w[i] and dis[u = to[i]] == -1) dis[u] = dis[x] + 1, q[++tl] = u;
  }
  return dis[T] != -1;
}
long long dfs(long long x, long long flow) {
  if (x == T) return flow;
  long long used = 0, u;
  for (long long &i = cur[x]; i; i = nxt[i])
    if (dis[u = to[i]] == dis[x] + 1 and w[i]) {
      long long v = dfs(u, min(flow - used, w[i]));
      w[i] -= v, w[i ^ 1] += v, used += v;
      if (used == flow) return flow;
    }
  return used;
}
inline long long dinic() {
  long long ans = 0;
  while (bfs()) memcpy(cur, head, sizeof cur), ans += dfs(S, INF);
  return ans;
}
inline long long id(long long x, long long y) {
  static long long tot, id[maxn][205];
  if (!id[x][y + 101]) id[x][y + 101] = ++tot;
  return id[x][y + 101];
}
inline void work() {
  n = read(), m = read();
  for (long long i = 1; i <= n; i++)
    f[i].a = read(), f[i].b = read(), f[i].c = read();
  for (long long i = 1; i <= n; i++) {
    l[i] = read(), r[i] = read();
    for (long long j = l[i]; j <= r[i] + 1; j++)
      addedge(id(i, j), id(i, j + 1), lim - f[i](j));
  }
  for (long long i = 1; i <= n; i++)
    addedge(S, id(i, l[i]), INF), addedge(id(i, r[i] + 1), T, INF);
  while (m--) {
    long long a = read(), b = read(), d = read();
    for (long long i = l[a]; i <= r[a] + 1; i++)
      if (i - d >= l[b] and i - d <= r[b] + 1)
        addedge(id(a, i), id(b, i - d), INF);
  }
  printf("%d\n", n * lim - dinic());
}
}  // namespace star
signed main() {
  star::work();
  return 0;
}