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 R(T &xx) {
  xx = 0;
  char ch = getchar();
  bool F = 0;
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') F = 1, ch = getchar();
  while (ch >= '0' && ch <= '9')
    xx = xx + xx + (xx << 3) + ch - 48, ch = getchar();
  if (F) xx = -xx;
}
template <class T>
inline void add(T &x, T &y) {
  for (x += y; x >= 1000000007; x -= 1000000007)
    ;
}
int n, m, b[55][233];
struct nod {
  int a, b, c, l, r;
  inline int _f(int x) { return a * x * x + b * x + c; }
} p[55];
struct edg {
  int b, v, nx;
} e[10111 << 4];
int es, H[10111];
int cnt[10111], low[10111], cur[10111], dis[10111], pre[10111];
inline void ins(int x, int y, int z) {
  e[es].b = y;
  e[es].v = z;
  e[es].nx = H[x];
  H[x] = es++;
  e[es].b = x;
  e[es].v = 0;
  e[es].nx = H[y];
  H[y] = es++;
}
int sap(int S, int T, int N) {
  for (int i = 0; i <= N; i++) cur[i] = H[i];
  memset(low, 0, sizeof(low));
  memset(cnt, 0, sizeof(cnt));
  memset(dis, 0, sizeof(dis));
  low[S] = (1 << 30);
  cnt[S] = N + 1;
  int flow = 0;
  for (int u = S, v, nw; dis[S] <= N;) {
    for (nw = cur[u]; ~nw; nw = e[nw].nx)
      if (e[nw].v && dis[u] == dis[e[nw].b] + 1) break;
    cur[u] = nw;
    if (~nw) {
      pre[v = e[nw].b] = nw;
      low[v] = min(low[u], e[nw].v);
      if ((u = v) == T)
        for (flow += low[T]; u ^ S; u = e[pre[u] ^ 1].b)
          e[pre[u]].v -= low[T], e[pre[u] ^ 1].v += low[T];
    } else {
      if (--cnt[dis[u]] == 0) break;
      cur[u] = H[u];
      dis[u] = N + 1;
      for (nw = H[u]; ~nw; nw = e[nw].nx)
        if (e[nw].v) dis[u] = min(dis[u], dis[e[nw].b] + 1);
      ++cnt[dis[u]];
      if (u ^ S) u = e[pre[u] ^ 1].b;
    }
  }
  return flow;
}
int main() {
  R(n);
  R(m);
  memset(H, -1, sizeof(H));
  for (int i = 1; i <= n; i++) R(p[i].a), R(p[i].b), R(p[i].c);
  int mx = 1e6, tt = 1, u, v, d;
  for (int i = 1; i <= n; i++) {
    R(p[i].l), R(p[i].r);
    for (int j = p[i].l - 1; j <= p[i].r; j++) b[i][j + 101] = ++tt;
    for (int j = p[i].l; j <= p[i].r; j++)
      ins(b[i][j + 100], b[i][j + 101], mx - p[i]._f(j));
  }
  int S = 1, T = ++tt;
  for (int i = 1; i <= n; i++)
    ins(S, b[i][p[i].l + 100], (1 << 30)),
        ins(b[i][p[i].r + 101], T, (1 << 30));
  for (int i = 1; i <= m; i++) {
    R(u), R(v), R(d);
    for (int j = p[u].l - 1; j <= p[u].r; j++)
      if (j - d >= p[v].l - 1 && j - d <= p[v].r)
        ins(b[u][j + 101], b[v][j - d + 101], (1 << 30));
  }
  printf("%d\n", mx * n - sap(S, T, T));
  return 0;
}