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 MAXN = 50 + 5, MAXM = 200 + 10, MAXT = MAXN * MAXM, INF = 0x3f3f3f3f;
struct Graph {
  int id, g, next;
} g[MAXT * MAXM * 2];
int n, m, S, T, tot, Tohka, l[MAXN], r[MAXN], a[MAXN], b[MAXN], c[MAXN],
    st[MAXT], v[MAXN];
int num[MAXT], dis[MAXT];
void Add(int first, int second, int G) {
  g[tot].id = second, g[tot].g = G, g[tot].next = st[first], st[first] = tot++;
  g[tot].id = first, g[tot].g = 0, g[tot].next = st[second], st[second] = tot++;
}
int C(int first, int second) {
  if (second < l[first]) second = l[first];
  if (second > r[first]) second = r[first] + 1;
  return v[first - 1] + second - l[first] + 1;
}
int calc(int id, int first) {
  return a[id] * ((first) * (first)) + b[id] * first + c[id];
}
void init() {
  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]), v[i] = v[i - 1] + r[i] - l[i] + 2;
  S = 0, T = v[n] + 1;
  memset(st, -1, sizeof(st));
  for (int i = 1; i <= n; ++i) {
    int first = l[i];
    for (int j = l[i]; j <= r[i]; ++j)
      if (calc(i, first) < calc(i, j)) first = j;
    Tohka += calc(i, first);
    Add(S, C(i, l[i]), INF);
    Add(C(i, r[i] + 1), T, INF);
    for (int j = l[i]; j <= r[i]; ++j)
      Add(C(i, j), C(i, j + 1), calc(i, first) - calc(i, j));
  }
  for (int i = 1; i <= m; ++i) {
    int a, b, d;
    scanf("%d%d%d", &a, &b, &d);
    for (int j = l[b]; j <= r[b]; ++j)
      if (j + d <= r[a]) Add(C(a, j + d + 1), C(b, j + 1), INF);
  }
}
int SAP(int first, int S, int T, int flow) {
  if (first == T) return flow;
  int res = flow;
  for (int i = st[first]; i != -1; i = g[i].next)
    if (g[i].g > 0 && dis[first] == dis[g[i].id] + 1) {
      int F = SAP(g[i].id, S, T, min(res, g[i].g));
      g[i].g -= F, g[i ^ 1].g += F, res -= F;
      if (res <= 0) return flow;
    }
  --num[dis[first]];
  if (!num[dis[first]]) dis[S] = T + 1;
  ++dis[first];
  ++num[dis[first]];
  return flow - res;
}
int Maxflow(int S, int T) {
  memset(num, 0, sizeof(num));
  memset(dis, 0, sizeof(dis));
  int Tohka = 0;
  for (num[0] = T + 1; dis[S] < T + 1; Tohka += SAP(S, S, T, INF))
    ;
  return Tohka;
}
void solve() { cout << Tohka - Maxflow(S, T) << endl; }
int main() {
  init();
  solve();
  fclose(stdin);
  fclose(stdout);
  return 0;
}