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;
int gap[11000], dis[11000], pre[11000], cur[11000];
int size, n, head[11000];
struct Node {
  int c, v, next;
  Node() {}
  Node(int _v, int _c, int _next) : v(_v), c(_c), next(_next) {}
} E[110000];
int sap(int s, int t) {
  memset(dis, 0, sizeof(dis));
  memset(gap, 0, sizeof(gap));
  for (int i = 1; i <= n; i++) cur[i] = head[i];
  int u = pre[s] = s, maxflow = 0, aug = -1;
  gap[0] = n;
  while (dis[s] < n) {
  loop:
    for (int &i = cur[u]; i != -1; i = E[i].next) {
      int v = E[i].v;
      if (E[i].c && dis[u] == dis[v] + 1) {
        if (aug == -1 || aug > E[i].c) aug = E[i].c;
        pre[v] = u;
        u = v;
        if (v == t) {
          maxflow += aug;
          for (u = pre[u]; v != s; v = u, u = pre[u]) {
            E[cur[u]].c -= aug;
            E[cur[u] ^ 1].c += aug;
          }
          aug = -1;
        }
        goto loop;
      }
    }
    int mindis = n;
    for (int i = head[u]; i != -1; i = E[i].next) {
      int v = E[i].v;
      if (E[i].c && mindis > dis[v]) {
        cur[u] = i;
        mindis = dis[v];
      }
    }
    if ((--gap[dis[u]]) == 0) break;
    gap[dis[u] = mindis + 1]++;
    u = pre[u];
  }
  return maxflow;
}
void add(int u, int v, int c, int cc = 0) {
  E[size] = Node(v, c, head[u]);
  head[u] = size++;
  E[size] = Node(u, cc, head[v]);
  head[v] = size++;
}
void init() {
  memset(head, -1, sizeof(head));
  size = 0;
}
int nn, mm;
int xi[55][3], L[55], R[55];
int res[110][3];
void read() {
  cin >> nn >> mm;
  for (int i = 0; i < nn; i++)
    for (int f = 0; f < 3; f++) scanf("%d", &xi[i][f]);
  for (int i = 0; i < nn; i++) scanf("%d%d", L + i, R + i);
  for (int i = 0; i < mm; i++) {
    for (int f = 0; f < 3; f++) scanf("%d", &res[i][f]);
    res[i][0]--;
    res[i][1]--;
  }
}
int getid(int a, int b) { return a * 205 + 1 + (b + 101); }
int maxx = 1000005;
int cal(int a, int b) { return xi[a][0] * b * b + xi[a][1] * b + xi[a][2]; }
int fun() {
  init();
  n = 10500;
  int s = n - 1, t = n;
  for (int i = 0; i < nn; i++) {
    add(s, getid(i, L[i] - 1), 0x3f3f3f3f);
    add(getid(i, R[i]), t, 0x3f3f3f3f);
    for (int f = L[i] - 1; f < R[i]; f++) {
      add(getid(i, f), getid(i, f + 1), maxx - cal(i, f + 1));
    }
  }
  for (int i = 0; i < mm; i++) {
    int u = res[i][0], v = res[i][1], d = res[i][2];
    for (int f = L[u] - 1; f <= R[u]; f++) {
      int oth = f - d;
      if (oth <= R[v] && oth >= L[v]) {
        add(getid(u, f), getid(v, oth), 0x3f3f3f3f);
      }
    }
  }
  int ans = sap(s, t);
  return maxx * nn - ans;
}
int main() {
  read();
  cout << fun() << endl;
  return 0;
}