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 maxm = 1e5 + 5;
const int maxn = 1e4 + 5;
const int inf = 1e9;
struct edge {
  int to, nex, w;
} e[maxm << 1];
int head[maxn], cur[maxn], cnt = 1;
int n, m, s, t;
void add(int a, int b, int c) {
  e[++cnt] = (edge){b, head[a], c};
  head[a] = cnt;
}
void addedge(int a, int b, int c) { add(a, b, c), add(b, a, 0); }
int dis[maxn];
int bfs() {
  memset(dis, 0, sizeof dis);
  queue<int> Q;
  dis[s] = 1;
  Q.push(s);
  for (int i = s; i <= t; ++i) cur[i] = head[i];
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int i = head[u]; i; i = e[i].nex) {
      int v = e[i].to;
      if (e[i].w && !dis[v]) {
        dis[v] = dis[u] + 1;
        Q.push(v);
      }
    }
  }
  return dis[t];
}
int dfs(int u, int in) {
  if (u == t) return in;
  int out = 0, tmp;
  for (int i = cur[u]; i; i = e[i].nex) {
    int v = e[i].to;
    cur[u] = i;
    if (e[i].w && dis[v] == dis[u] + 1 && (tmp = dfs(v, min(in, e[i].w)))) {
      e[i].w -= tmp;
      e[i ^ 1].w += tmp;
      in -= tmp;
      out += tmp;
      if (!in) break;
    }
  }
  if (!out) dis[u] = 0;
  return out;
}
int a[maxn], b[maxn], c[maxn];
int l[maxn], r[maxn];
int num;
int id[500][500];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> a[i] >> b[i] >> c[i];
  for (int i = 1; i <= n; ++i) cin >> l[i] >> r[i];
  for (int i = 1; i <= n; ++i) {
    for (int j = l[i]; j <= r[i] + 1; ++j) id[i][j + 100] = ++num;
    for (int j = l[i]; j <= r[i]; ++j)
      addedge(id[i][j + 100], id[i][j + 100 + 1],
              inf - (a[i] * j * j + b[i] * j + c[i]));
  }
  s = 0, t = ++num;
  for (int i = 1; i <= n; ++i) {
    addedge(s, id[i][l[i] + 100], 2 * inf);
    addedge(id[i][r[i] + 100 + 1], t, 2 * inf);
  }
  for (int i = 1; i <= m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    for (int i = max(l[u] - w, l[v]); i <= min(r[u] - w, r[v]) + 1; ++i)
      addedge(id[u][i + 100 + w], id[v][i + 100], 2 * inf);
  }
  long long ans = 1ll * n * inf;
  while (bfs()) {
    ans -= dfs(s, 2 * inf);
  }
  cout << ans << '\n';
  return 0;
}