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;
int n, m, a[50 + 1], b[50 + 1], c[50 + 1], l[50 + 1], r[50 + 1];
int f(int i, int x) { return a[i] * x * x + b[i] * x + c[i]; }
struct edge {
  int next, node;
  long long w;
} e[100000 << 1 | 1];
int head[20000 + 1], tot = 1;
void addedge(int a, int b, long long w) {
  e[++tot].next = head[a];
  head[a] = tot, e[tot].node = b, e[tot].w = w;
  e[++tot].next = head[b];
  head[b] = tot, e[tot].node = a, e[tot].w = 0;
}
int S, T, cnt = 0, st[50 + 1], d[20000 + 1], q[20000 + 1], M = 0;
int node(int x, int v) { return st[x] + v - l[x]; }
bool bfs() {
  int h = 0, t = 0;
  for (int i = S; i <= T; ++i) d[i] = 0;
  d[S] = 1, q[t++] = S;
  while (h < t) {
    int cur = q[h++];
    for (int i = head[cur]; i; i = e[i].next) {
      if (!e[i].w) continue;
      int node = e[i].node;
      if (d[node]) continue;
      d[node] = d[cur] + 1;
      q[t++] = node;
    }
  }
  return d[T] != 0;
}
long long dfs(int x, long long inflow) {
  if (x == T) return inflow;
  long long ret = inflow, flow;
  for (int i = head[x]; i; i = e[i].next) {
    if (!e[i].w) continue;
    int node = e[i].node;
    if (d[node] != d[x] + 1) continue;
    flow = dfs(node, min(e[i].w, ret));
    if (!flow) continue;
    ret -= flow, e[i].w -= flow, e[i ^ 1].w += flow;
    if (!ret) break;
  }
  if (ret == inflow) d[x] = -1;
  return inflow - ret;
}
long long maxFlow() {
  long long ret = 0LL;
  while (bfs()) ret += dfs(S, (long long)n * 87654321 * 2);
  return ret;
}
int main(int argc, char *argv[]) {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> a[i] >> b[i] >> c[i];
  S = 0, T = 1;
  for (int i = 1; i <= n; ++i) {
    cin >> l[i] >> r[i];
    st[i] = T;
    T += r[i] - l[i] + 1 + 1;
    for (int j = l[i]; j <= r[i]; ++j) M = max(M, f(i, j) + 1);
  }
  M = 210000;
  for (int i = 1; i <= n; ++i) {
    addedge(S, st[i], 87654321);
    for (int j = l[i]; j <= r[i]; ++j)
      addedge(node(i, j), node(i, j) + 1, M - f(i, j));
    addedge(node(i, r[i]) + 1, T, 87654321);
  }
  for (int i = 1; i <= m; ++i) {
    int x, y, w;
    cin >> x >> y >> w;
    for (int j = l[x]; j <= r[x]; ++j)
      if (j - w >= l[y] && j - w <= r[y] + 1)
        addedge(node(x, j), node(y, j - w), 87654321);
  }
  long long ans = maxFlow();
  cerr << ans << endl;
  if (ans >= 87654321)
    cout << "Invalid" << endl;
  else {
    ans = (long long)M * n - ans;
    cout << ans << endl;
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}