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 = ~0U >> 1;
template <class Flow>
struct Maxflow {
  static const Flow INF = ~0U >> 1;
  struct Edge {
    int t;
    Flow c;
    Edge *n, *r;
    Edge(int _t, Flow _c, Edge* _n) : t(_t), c(_c), n(_n) {}
  };
  vector<Edge*> E;
  int addV() {
    E.push_back((Edge*)0);
    return E.size() - 1;
  }
  void clear() { E.clear(); }
  Edge* makeEdge(int s, int t, Flow c) { return E[s] = new Edge(t, c, E[s]); }
  void addEdge(int s, int t, Flow c) {
    Edge *e1 = makeEdge(s, t, c), *e2 = makeEdge(t, s, 0);
    e1->r = e2, e2->r = e1;
  }
  int calcMaxFlow(int vs, int vt) {
    int nV = E.size();
    Flow totalFlow = 0;
    vector<Flow> am(nV, 0);
    vector<int> h(nV, 0), cnt(nV + 1, 0);
    vector<Edge*> prev(nV, (Edge*)0), cur(nV, (Edge*)0);
    cnt[0] = nV;
    int u = vs;
    Edge* e;
    am[u] = INF;
    while (h[vs] < nV) {
      for (e = cur[u]; e; e = e->n)
        if (e->c > 0 && h[u] == h[e->t] + 1) break;
      if (e) {
        int v = e->t;
        cur[u] = prev[v] = e;
        am[v] = min(am[u], e->c);
        u = v;
        if (u == vt) {
          Flow by = am[u];
          while (u != vs) {
            prev[u]->c -= by;
            prev[u]->r->c += by;
            u = prev[u]->r->t;
          }
          totalFlow += by;
          am[u] = INF;
        }
      } else {
        if (!--cnt[h[u]]) break;
        h[u] = nV;
        for (e = E[u]; e; e = e->n)
          if (e->c > 0 && h[e->t] + 1 < h[u]) {
            h[u] = h[e->t] + 1;
            cur[u] = e;
          }
        ++cnt[h[u]];
        if (u != vs) u = prev[u]->r->t;
      }
    }
    return totalFlow;
  }
  ~Maxflow() {
    for (int i = 0; i < E.size(); ++i) {
      for (Edge* e = E[i]; e;) {
        Edge* ne = e->n;
        delete e;
        e = ne;
      }
    }
  }
};
Maxflow<int> net;
int n, m;
const int MAX_N = 50 + 1, MAX_M = 100 + 1;
int a[MAX_N], b[MAX_N], c[MAX_N];
const int BIG = int(3e5);
const int MX = 110;
vector<int> id[MAX_N];
int l[MAX_M], r[MAX_M];
int vs, vt;
int eval(int i, int x) { return (a[i] * x + b[i]) * x + c[i]; }
int get(int u, int i) {
  if (i < -MX || i > MX) return -1;
  return id[u][i + MX];
}
void addEdge(int u, int i, int v, int j) {
  int x = get(u, i), y = get(v, j);
  if (x == -1 || y == -1) return;
  net.addEdge(x, y, INF);
}
void constraint(int u, int v, int d) {
  for (int i = -MX; i <= MX; ++i) {
    addEdge(u, i, v, i - d);
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> a[i] >> b[i] >> c[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> l[i] >> r[i];
  }
  vs = net.addV(), vt = net.addV();
  for (int me = 0; me < n; ++me) {
    int tot = MX * 2 + 2;
    vector<int>& v = id[me];
    for (int j = 0; j < tot; ++j) {
      v.push_back(net.addV());
    }
    net.addEdge(vs, v[0], INF);
    for (int i = -MX; i <= MX; ++i) {
      if (i >= l[me] && i <= r[me])
        net.addEdge(v[i + MX], v[i + MX + 1], BIG - eval(me, i));
      else
        net.addEdge(v[i + MX], v[i + MX + 1], INF);
    }
    net.addEdge(v.back(), vt, INF);
  }
  while (m--) {
    int u, v, d;
    cin >> u >> v >> d;
    --u, --v;
    constraint(u, v, d);
  }
  int ans = BIG * n - net.calcMaxFlow(vs, vt);
  cout << ans << endl;
}