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 read(T& x) {
  bool fu = 0;
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') fu = 1, c = getchar();
  for (x = 0; c > 32; c = getchar()) x = x * 10 + c - '0';
  if (fu) x = -x;
};
template <class T>
inline void read(T& x, T& y) {
  read(x);
  read(y);
}
template <class T>
inline void read(T& x, T& y, T& z) {
  read(x);
  read(y);
  read(z);
}
inline char getc() {
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  return c;
}
const int V = 20010;
const int E = 5000010;
const int inf = int(1e9);
int S, T, N, edge, e[E], b[E], c[E], fir[V], last[V];
void clear() {
  edge = 0;
  memset(fir, -1, sizeof(fir));
  memset(b, -1, sizeof(b));
}
void add2(int x, int y, int z) {
  e[edge] = y;
  c[edge] = z;
  if (fir[x] == -1)
    fir[x] = edge;
  else
    b[last[x]] = edge;
  last[x] = edge++;
}
void add(int x, int y, int z) {
  add2(x, y, z);
  add2(y, x, 0);
}
int vh[V], h[V], cur[V], pre[V];
int sap() {
  int i, j, k, l, p, flow, minh, ans = 0;
  memset(h, 0, sizeof(h));
  memset(vh, 0, sizeof(vh));
  vh[0] = N;
  memcpy(cur, fir, sizeof(fir));
  memset(pre, 0, sizeof(pre));
  pre[S] = S;
  for (i = S; h[S] < N;) {
    if (i == T) {
      flow = inf;
      for (p = S; p != T; p = e[cur[p]]) flow = min(flow, c[cur[p]]);
      ans += flow;
      for (p = S; p != T; p = e[cur[p]])
        c[cur[p]] -= flow, c[cur[p] ^ 1] += flow;
      i = S;
    }
    for (k = cur[i]; k >= 0; k = b[k])
      if (c[k] && h[e[k]] + 1 == h[i]) {
        cur[i] = k;
        pre[e[k]] = i;
        i = e[k];
        break;
      }
    if (k < 0) {
      if (--vh[h[i]] == 0) break;
      cur[i] = fir[i];
      minh = N;
      for (k = cur[i]; k >= 0; k = b[k])
        if (c[k]) minh = min(minh, h[e[k]] + 1);
      ++vh[h[i] = minh];
      i = pre[i];
    }
  }
  return ans;
}
const int D = 101;
int n, m, i, j, k, l, p, x, y, z, op, tot;
int A[51], B[51], C[51], L[51], R[51];
int n0[51][210];
int main() {
  read(n, m);
  for (i = 1; i <= n; i++) read(A[i], B[i], C[i]);
  for (i = 1; i <= n; i++) read(L[i], R[i]);
  for (i = 1; i <= n; i++)
    for (j = L[i]; j <= R[i] + 1; j++) n0[i][j + D] = ++tot;
  S = V - 2;
  N = T = V - 1;
  clear();
  op = int(1e7);
  for (i = 1; i <= n; i++) {
    add(S, n0[i][L[i] + D], inf);
    add(n0[i][R[i] + 1 + D], T, inf);
    for (j = L[i]; j <= R[i]; j++)
      add(n0[i][j + D], n0[i][j + 1 + D],
          op - (A[i] * j * j + B[i] * j + C[i]));
  }
  for (i = 1; i <= m; i++) {
    read(x, y, z);
    for (j = L[x]; j <= R[x]; j++) {
      int j2 = j - z;
      if (j2 > R[y]) {
        add(n0[x][j + D], n0[x][j + 1 + D], inf);
        continue;
      }
      if (j2 >= L[y]) add(n0[x][j + D], n0[y][j2 + D], inf);
    }
  }
  cout << op * n - sap() << endl;
  return 0;
}