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>
namespace MF {
long long n, m, s, t;
long long hd[20005], _hd;
struct edge {
  long long v, f, nxt;
} e[100005 << 1];
inline void addedge(long long u, long long v, long long f) {
  e[++_hd] = (edge){v, f, hd[u]};
  hd[u] = _hd;
  e[++_hd] = (edge){u, 0, hd[v]};
  hd[v] = _hd;
}
long long cur[20005], dis[20005];
std::queue<long long> q;
inline bool bfs(long long s, long long t) {
  for (long long i = 1; i <= n; i++) cur[i] = hd[i];
  for (long long i = 1; i <= n; i++) dis[i] = 0x3f3f3f3f3f;
  dis[s] = 0;
  q.push(s);
  while (!q.empty()) {
    long long u = q.front();
    q.pop();
    for (long long i = hd[u]; i; i = e[i].nxt) {
      long long v = e[i].v, f = e[i].f;
      if (f && dis[v] > dis[u] + 1) {
        dis[v] = dis[u] + 1;
        q.push(v);
      }
    }
  }
  return dis[t] < 0x3f3f3f3f3f;
}
inline long long dfs(long long u, long long lmt, long long t) {
  if (u == t || !lmt) return lmt;
  long long flw = 0;
  for (long long i = cur[u]; i; i = e[i].nxt) {
    cur[u] = i;
    long long v = e[i].v, f = e[i].f;
    if (dis[v] != dis[u] + 1) continue;
    f = dfs(v, std::min(lmt, f), t);
    if (f) {
      e[i].f -= f;
      e[i ^ 1].f += f;
      flw += f;
      lmt -= f;
      if (!lmt) break;
    }
  }
  return flw;
}
inline long long dinic(long long s, long long t) {
  long long res = 0;
  while (bfs(s, t)) res += dfs(s, 0x3f3f3f3f3f, t);
  return res;
}
long long f[20005];
inline void init(long long n_, long long s_, long long t_) {
  for (long long i = 1; i <= n; i++) hd[i] = 0;
  _hd = 1;
  n = n_, s = s_, t = t_;
}
inline long long sol() { return dinic(s, t); }
}  // namespace MF
using MF::addedge;
using MF::init;
using MF::sol;
long long n, m, a[20005], b[20005], c[20005], l[20005], r[20005];
inline long long num(long long x, long long y) { return (x)*300 + y + 150; }
long long s, t;
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 = num(n, 101) + 1, t = num(n, 101) + 2;
  init(num(n, 101) + 2, s, t);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld%lld", &l[i], &r[i]);
    addedge(s, num(i, l[i]), 0x3f3f3f3f3f);
    for (long long j = l[i]; j <= r[i]; j++)
      addedge(num(i, j), num(i, j + 1),
              (0x3f3f3f3f3f >> 7) - (a[i] * j * j + b[i] * j + c[i]));
    addedge(num(i, r[i] + 1), t, 0x3f3f3f3f3f);
  }
  for (long long i = 1; i <= m; i++) {
    long long u, v, d;
    scanf("%lld%lld%lld", &u, &v, &d);
    for (long long j = l[u]; j <= r[u]; j++)
      if (j - d >= l[v] & j - d <= r[v] + 1)
        addedge(num(u, j), num(v, j - d), 0x3f3f3f3f3f);
  }
  printf("%lld\n", n * (0x3f3f3f3f3f >> 7) - sol());
}