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;
const int M(10000000);
struct LnkNode {
  int v, w, nxt;
} edge[(55 * 209 + 105 * 209) * 2];
int etot = 1, fst[55 * 209], cur[55 * 209];
void addedge_(int u, int v, int w) {
  ++etot;
  edge[etot].v = v;
  edge[etot].w = w;
  edge[etot].nxt = fst[u];
  fst[u] = etot;
}
void addedge(int u, int v, int w) { addedge_(u, v, w), addedge_(v, u, 0); }
int n, m, ncnt, nd[55][209], a[55], b[55], c[55], l[55], r[55];
int d[55 * 209], q[55 * 209], Head, Tail;
bool bfs(int s, int t) {
  memset(d + 1, 0x3f, (ncnt + 2) << 2);
  Head = Tail = 0;
  d[q[Tail++] = s] = 0;
  while (Head ^ Tail) {
    int &x = q[Head++];
    for (int i = fst[x]; i; i = edge[i].nxt) {
      int &y = edge[i].v, &w = edge[i].w;
      if (w && d[y] > d[x] + 1) d[q[Tail++] = y] = d[x] + 1;
    }
  }
  return d[t] != 0x3f3f3f3f;
}
int dfs(int x, int t, int f) {
  if (x == t) return f;
  int r = f;
  for (int &i = cur[x]; i; i = edge[i].nxt) {
    int &y = edge[i].v, &w = edge[i].w;
    if (w && d[y] == d[x] + 1) {
      int k = dfs(y, t, min(r, w));
      w -= k, edge[i ^ 1].w += k;
      if (!(r -= k)) break;
    }
  }
  return f - r;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d%d%d", a + i, b + i, c + i);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", l + i, r + i);
    for (int j = l[i] - 1; j <= r[i]; ++j) nd[i][j + 101] = ++ncnt;
  }
  for (int i = 1; i <= n; ++i) {
    addedge(ncnt + 1, nd[i][l[i] - 1 + 101], 0x7fffffff);
    for (int j = l[i]; j <= r[i]; ++j)
      addedge(nd[i][j - 1 + 101], nd[i][j + 101],
              M - (a[i] * j * j + b[i] * j + c[i]));
    addedge(nd[i][r[i] + 101], ncnt + 2, 0x7fffffff);
  }
  for (int i = 1, u, v, d; i <= m; ++i) {
    scanf("%d%d%d", &u, &v, &d);
    for (int j = l[u] - 1; j <= r[u]; ++j)
      if (l[v] - 1 <= j - d && j - d <= r[v])
        addedge(nd[u][j + 101], nd[v][j - d + 101], 0x7fffffff);
      else if (j - d > r[v])
        addedge(nd[u][j + 101], ncnt + 2, 0x7fffffff);
  }
  long long f, mf = 0;
  while (bfs(ncnt + 1, ncnt + 2)) {
    memcpy(cur + 1, fst + 1, (ncnt + 2) << 2);
    while ((f = dfs(ncnt + 1, ncnt + 2, 0x7fffffff))) mf += f;
  }
  printf("%lld\n", (long long)M * n - mf);
  return 0;
}