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;
template <class T>
inline void chkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void chkmax(T &a, T b) {
  if (a < b) a = b;
}
namespace Dinic {
int en;
int fst[11000], prv[11000], q[11000];
int nxt[110000], dst[110000], cap[110000];
void init() {
  en = 0;
  memset(fst, -1, sizeof fst);
}
inline void insert(int u, int v, int c) {
  nxt[en] = fst[u];
  fst[u] = en;
  dst[en] = v;
  cap[en++] = c;
  nxt[en] = fst[v];
  fst[v] = en;
  dst[en] = u;
  cap[en++] = 0;
}
int maxFlow(int S, int T, int n) {
  int rlt(0), hd, tl, u, v, i, j, df;
  while (1) {
    memset(prv, -1, sizeof(int) * n);
    tl = 0;
    q[tl++] = S;
    prv[S] = -2;
    int tot = 0;
    for (hd = 0; hd < tl; hd++) {
      for (i = fst[q[hd]]; i != -1; i = nxt[i]) {
        v = dst[i];
        if (prv[v] == -1 && cap[i]) {
          prv[v] = i ^ 1;
          q[tl++] = v;
        }
      }
    }
    if (prv[T] == -1) break;
    for (i = fst[T]; i != -1; i = nxt[i]) {
      u = dst[i];
      if (prv[u] == -1) continue;
      df = cap[i ^ 1];
      for (j = prv[u]; j >= 0 && df; j = prv[dst[j]]) {
        chkmin(df, cap[j ^ 1]);
      }
      if (df == 0) continue;
      rlt += df;
      cap[i ^ 1] -= df;
      cap[i] += df;
      for (j = prv[u]; j >= 0; j = prv[dst[j]]) {
        cap[j ^ 1] -= df;
        cap[j] += df;
      }
    }
  }
  return rlt;
}
};  // namespace Dinic
int n, m;
const int MX = 55;
int a[MX], b[MX], c[11000], L[MX], R[MX];
inline int calc(int i, int x) {
  return 1000000 - (a[i] * x * x + b[i] * x + c[i]);
}
int main() {
  int i, j, k, u, v, S, T, N, d;
  scanf("%d%d", &n, &m);
  for (i = 0; i < n; i++) scanf("%d%d%d", a + i, b + i, c + i);
  S = 205 * n;
  T = 205 * n + 1, N = 205 * n + 2;
  Dinic::init();
  for (i = 0; i < n; i++) {
    scanf("%d%d", L + i, R + i);
    for (j = 0; j <= R[i] - L[i]; j++) {
      Dinic::insert(i * 205 + j, i * 205 + j + 1, calc(i, L[i] + j));
    }
    Dinic::insert(S, i * 205, int(1e9));
    Dinic::insert(i * 205 + R[i] - L[i] + 1, T, int(1e9));
  }
  for (i = 0; i < m; i++) {
    scanf("%d%d%d", &u, &v, &d);
    u--, v--;
    for (j = L[u]; j <= R[u]; j++) {
      if (j - d > R[v]) {
        Dinic::insert(u * 205 + j - L[u], v * 205 + R[v] - L[v] + 1, int(1e9));
        continue;
      }
      if (j - d < L[v]) continue;
      if (j - d >= L[v] && j - d <= R[v]) {
        Dinic::insert(u * 205 + j - L[u], v * 205 + j - d - L[v], int(1e9));
      }
    }
  }
  printf("%d\n", -Dinic::maxFlow(S, T, N) + 1000000 * n);
  return 0;
}