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;
const int BIG = 4000000;
const int INF = 0x3f3f3f3f;
template <int maxn, int maxm>
struct ISAP {
  struct Node {
    int v, c, next;
  } Edge[maxm * 2 + 10];
  int adj[maxn + 10], ecnt;
  int gap[maxn + 10], d[maxn + 10];
  int n, m, s, t;
  void addedge(int u, int v, int c) {
    Node &e = Edge[ecnt];
    e.v = v;
    e.c = c;
    e.next = adj[u];
    adj[u] = ecnt++;
  }
  void add(int u, int v, int c) {
    addedge(u, v, c);
    addedge(v, u, 0);
  }
  void init() {
    memset(adj, -1, sizeof(adj));
    ecnt = 0;
  }
  void init_dis() {
    memset(d, -1, sizeof(d));
    memset(gap, 0, sizeof(gap));
    queue<int> q;
    d[t] = 0;
    q.push(t);
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      n++;
      gap[d[u]]++;
      for (int i = adj[u]; ~i; i = Edge[i].next) {
        int v = Edge[i].v;
        if (d[v] == -1) d[v] = d[u] + 1, q.push(v);
      }
    }
  }
  int aug(int u, int inc) {
    int Inc = 0, mindis = n - 1;
    if (u == t) return inc;
    for (int i = adj[u]; ~i; i = Edge[i].next) {
      int v = Edge[i].v, c = Edge[i].c;
      if (c) {
        if (d[v] == d[u] - 1) {
          int del = min(c, inc - Inc);
          del = aug(v, del);
          Inc += del;
          Edge[i].c -= del;
          Edge[i ^ 1].c += del;
          if (inc == Inc) return Inc;
          if (d[s] >= n) return Inc;
        }
        mindis = min(mindis, d[v]);
      }
    }
    if (!Inc) {
      gap[d[u]]--;
      if (gap[d[u]] == 0) d[s] = n;
      d[u] = mindis + 1;
      gap[d[u]]++;
    }
    return Inc;
  }
  int Maxflow(int _s, int _t) {
    s = _s;
    t = _t;
    n = 0;
    int Flow = 0;
    init_dis();
    while (d[s] < n) Flow += aug(s, INF);
    return Flow;
  }
};
ISAP<20000, 2000000> sap;
int a[MAXN + 10], b[MAXN + 10], c[MAXN + 10], l[MAXN + 10], r[MAXN + 10];
int n, m, S, T;
int f(int i, int x) { return a[i] * x * x + b[i] * x + c[i]; }
inline int idx(int x, int num) {
  if (T && num < l[x]) return S;
  return (x - 1) * (300) + num + 101;
}
int main() {
  sap.init();
  scanf("%d%d", &n, &m);
  S = idx(n + 1, -100);
  T = idx(n + 1, -99);
  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 i = 1; i <= n; i++) {
    sap.add(S, idx(i, l[i]), BIG - f(i, l[i]));
    sap.add(idx(i, r[i]), T, INF);
    for (int j = l[i] + 1; j <= r[i]; j++)
      sap.add(idx(i, j - 1), idx(i, j), BIG - f(i, j));
  }
  for (int i = 1; i <= m; i++) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    for (int j = l[u]; j <= r[u] + 1; j++) {
      int t = j - d - 1;
      if (l[v] <= t && t <= r[v]) sap.add(idx(u, j - 1), idx(v, t), INF);
    }
  }
  int ans = sap.Maxflow(S, T);
  printf("%d\n", -ans + BIG * n);
}