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 INF = (int)1e9 + 7;
const int MAXV = (int)11e6;
class DinicFlow {
 private:
  int n, m;
  vector<int> dist, head, q, work;
  vector<int> next, capa, flow, point;

 public:
  DinicFlow() { n = 0; }
  DinicFlow(int n) {
    this->n = n;
    this->m = 0;
    dist.assign(n + 7, 0);
    head.assign(n + 7, -1);
    q.assign(n + 7, 0);
    work.assign(n + 7, 0);
  }
  void addedge(int u, int v, int c1, int c2) {
    point.push_back(v);
    capa.push_back(c1);
    flow.push_back(0);
    next.push_back(head[u]);
    head[u] = m++;
    point.push_back(u);
    capa.push_back(c2);
    flow.push_back(0);
    next.push_back(head[v]);
    head[v] = m++;
  }
  bool bfs(int s, int t) {
    for (int i = (1), _b = (n); i <= _b; i = i + 1) dist[i] = -1;
    int sz = 0;
    q[sz++] = s;
    dist[s] = 0;
    for (int x = 0; x < sz; x = x + 1) {
      int u = q[x];
      for (int i = head[u]; i >= 0; i = next[i])
        if (dist[point[i]] < 0 && flow[i] < capa[i]) {
          dist[point[i]] = dist[u] + 1;
          q[sz++] = point[i];
        }
    }
    return (dist[t] >= 0);
  }
  int dfs(int s, int t, int f) {
    if (s == t) return (f);
    for (int &i = work[s]; i >= 0; i = next[i])
      if (dist[point[i]] == dist[s] + 1 && flow[i] < capa[i]) {
        int d = dfs(point[i], t, min(f, capa[i] - flow[i]));
        if (d > 0) {
          flow[i] += d;
          flow[i ^ 1] -= d;
          return (d);
        }
      }
    return (0);
  }
  int maxflow(int s, int t) {
    int totflow = 0;
    while (bfs(s, t)) {
      for (int i = (1), _b = (n); i <= _b; i = i + 1) work[i] = head[i];
      while (true) {
        int d = dfs(s, t, INF);
        if (d == 0) break;
        totflow += d;
      }
    }
    return (totflow);
  }
};
struct var {
  int a, b, c, l, r;
  var() { a = b = c = l = r = 0; }
  int val(int x) const { return (a * x * x + b * x + c); }
};
inline int nextInt(void) {
  int x;
  scanf("%d", &x);
  return (x);
}
var v[55];
DinicFlow DF;
int n, m, s[55];
inline int nodeid(int x, int k) { return (s[x - 1] + k - v[x].l + 2); }
void process(void) {
  n = nextInt();
  m = nextInt();
  for (int i = (1), _b = (n); i <= _b; i = i + 1) {
    v[i].a = nextInt();
    v[i].b = nextInt();
    v[i].c = nextInt();
  }
  for (int i = (1), _b = (n); i <= _b; i = i + 1) {
    v[i].l = nextInt();
    v[i].r = nextInt();
    s[i] = s[i - 1] + v[i].r - v[i].l + 2;
  }
  DF = DinicFlow(s[n] + 2);
  int src = s[n] + 1;
  int snk = s[n] + 2;
  for (int i = (1), _b = (n); i <= _b; i = i + 1) {
    DF.addedge(src, nodeid(i, v[i].l - 1), INF, 0);
    DF.addedge(nodeid(i, v[i].r), snk, INF, 0);
    for (int j = (v[i].l - 1), _b = (v[i].r - 1); j <= _b; j = j + 1)
      DF.addedge(nodeid(i, j), nodeid(i, j + 1), MAXV - v[i].val(j + 1), 0);
  }
  for (int zz = 0, _n = (m); zz < _n; zz = zz + 1) {
    int x = nextInt();
    int y = nextInt();
    int d = nextInt();
    for (int i = (v[x].l - 1), _b = (v[x].r); i <= _b; i = i + 1) {
      int t = i - d;
      if (v[y].l - 1 <= t && t <= v[y].r)
        DF.addedge(nodeid(x, i), nodeid(y, t), INF, 0);
    }
  }
  printf("%d", MAXV * n - DF.maxflow(src, snk));
}
int main(void) {
  process();
  return 0;
}