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;
namespace Dinic {
const int MAXV = 1000000;
const int MAXE = 1000000;
const int oo = 1000000000;
int V, E;
int last[MAXV], dist[MAXV], curr[MAXV];
int next[MAXE], cap[MAXE], adj[MAXE];
void init(int n) {
  V = n;
  E = 0;
  for (int i = (0); i < (V); ++i) last[i] = -1;
}
void edge(int x, int y, int c1, int c2) {
  adj[E] = y;
  cap[E] = c1;
  next[E] = last[x];
  last[x] = E++;
  adj[E] = x;
  cap[E] = c2;
  next[E] = last[y];
  last[y] = E++;
}
int push(int x, int sink, int flow) {
  if (x == sink) return flow;
  for (int &e = curr[x]; e != -1; e = next[e]) {
    int y = adj[e];
    if (cap[e] && dist[x] + 1 == dist[y])
      if (int f = push(y, sink, min(flow, cap[e])))
        return cap[e] -= f, cap[e ^ 1] += f, f;
  }
  return 0;
}
int run(int src, int sink) {
  int ret = 0;
  for (;;) {
    for (int i = (0); i < (V); ++i) curr[i] = last[i];
    for (int i = (0); i < (V); ++i) dist[i] = -1;
    queue<int> Q;
    Q.push(src), dist[src] = 0;
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (int e = last[x]; e != -1; e = next[e]) {
        int y = adj[e];
        if (cap[e] && dist[y] == -1) Q.push(y), dist[y] = dist[x] + 1;
      }
    }
    if (dist[sink] == -1) break;
    while (int f = push(src, sink, oo)) ret += f;
  }
  return ret;
}
}  // namespace Dinic
const int oo = 1e9;
const int T = 1e6;
int N, M;
int a[50], b[50], c[50], l[50], r[50];
int main(void) {
  scanf("%d%d", &N, &M);
  for (int i = (0); i < (N); ++i) scanf("%d%d%d", a + i, b + i, c + i);
  for (int i = (0); i < (N); ++i) scanf("%d%d", l + i, r + i);
  int V = 222 * N;
  int src = V++;
  int sink = V++;
  Dinic::init(V);
  auto get = [src, sink](int i, int x) {
    if (x < l[i]) return src;
    if (x > r[i]) return sink;
    return 222 * i + x + 100;
  };
  auto output = [](int i, int x) {
    assert(l[i] <= x && x <= r[i]);
    return a[i] * x * x + b[i] * x + c[i];
  };
  for (int i = (0); i < (N); ++i) {
    Dinic::edge(src, get(i, l[i]), oo, 0);
    Dinic::edge(get(i, r[i]), sink, T - output(i, r[i]), 0);
    for (int x = (l[i]); x < (r[i]); ++x) {
      Dinic::edge(get(i, x), get(i, x + 1), T - output(i, x), 0);
    }
  }
  for (int i = (0); i < (M); ++i) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    --u, --v;
    d = -d;
    for (int x = (l[u]); x < (r[u] + 1); ++x)
      Dinic::edge(get(u, x), get(v, x + d), oo, 0);
  }
  printf("%d\n", N * T - Dinic::run(src, sink));
  return 0;
}