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>
long long lim = 1e8, inf = 0x3f3f3f3f3f3f3f3fll;
long long n, m, head[100010], to[200010], nxt[1200010], w[200010], cnt = 1, idx;
long long id[210][210], a[110], b[110], c[110], l[110], r[110], S, T;
long long getans(long long i, long long const &x) {
  return a[i] * x * x + b[i] * x + c[i];
}
void add_(long long const &x, long long const &y, long long const &z) {
  to[++cnt] = y, nxt[cnt] = head[x], w[cnt] = z, head[x] = cnt;
}
void add(long long const &x, long long const &y, long long const &z) {
  add_(x, y, z), add_(y, x, 0);
}
long long cur[100010], d[100010];
bool bfs() {
  for (long long i = 1; i <= idx; i++) d[i] = -1, cur[i] = head[i];
  static long long que[100010], *hd, *tl;
  hd = tl = que, *tl++ = S;
  d[S] = 0;
  while (hd != tl) {
    long long x = *hd++;
    for (long long i = head[x]; i; i = nxt[i])
      if (!~d[to[i]] && w[i]) d[to[i]] = d[x] + 1, *tl++ = to[i];
  }
  return d[T] != -1;
}
long long dfs(long long const &x, long long const &flow) {
  if (x == T) return flow;
  long long used = 0;
  for (long long &i = cur[x]; i; i = nxt[i])
    if (d[to[i]] == d[x] + 1) {
      long long v = dfs(to[i], std::min(w[i], flow - used));
      w[i] -= v;
      w[i ^ 1] += v;
      used += v;
      if (used == flow) return used;
    }
  return used;
}
long long dinic() {
  long long flow = 0;
  while (bfs()) flow += dfs(S, inf);
  return flow;
}
signed main() {
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) scanf("%lld%lld%lld", a + i, b + i, c + i);
  S = ++idx, T = ++idx;
  for (long long i = 1; i <= n; i++) {
    scanf("%lld%lld", l + i, r + i);
    for (long long j = l[i] + 100; j <= r[i] + 101; j++) {
      id[i][j] = ++idx;
      if (j != l[i] + 100)
        add(id[i][j - 1], id[i][j], lim - getans(i, j - 1 - 100));
      else
        add(S, id[i][j], inf);
    }
  }
  for (long long i = 1; i <= n; i++) add(id[i][r[i] + 101], T, inf);
  for (long long i = 1, u, v, d; i <= m; i++) {
    scanf("%lld%lld%lld", &u, &v, &d);
    for (long long j = l[u]; j <= r[u] + 1; j++)
      if (l[v] <= j - d && j - d <= r[v] + 1)
        add(id[u][j + 100], id[v][j - d + 100], inf);
  }
  printf("%lld\n", n * lim - dinic());
  return 0;
}