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 nmax = 12000;
const long long inf = 1e18, MX = 1e12;
struct edge {
  int from, to;
  long long flow;
};
vector<int> adj[nmax];
vector<edge> all;
int n, m;
void add_edge(int from, int to, long long flow) {
  edge current;
  current.from = from;
  current.to = to;
  current.flow = flow;
  adj[from].push_back(all.size());
  all.push_back(current);
  swap(current.from, current.to);
  current.flow = 0;
  adj[to].push_back(all.size());
  all.push_back(current);
}
int source, sink;
long long bfs_dist[nmax];
queue<pair<int, int> > active, idle;
void bfs() {
  for (int i = source; i <= sink; i++) bfs_dist[i] = inf;
  active = idle;
  active.push({source, 0});
  while (active.size()) {
    pair<int, int> current = active.front();
    active.pop();
    if (bfs_dist[current.first] != inf) continue;
    bfs_dist[current.first] = current.second;
    if (current.first == sink) return;
    for (auto k : adj[current.first])
      if (all[k].flow && bfs_dist[all[k].to] == inf)
        active.push({all[k].to, current.second + 1});
  }
}
bool been[nmax];
long long dfs(int node, long long cap) {
  if (node == sink) return cap;
  if (been[node]) return 0;
  been[node] = 1;
  for (auto k : adj[node]) {
    if (bfs_dist[node] < bfs_dist[all[k].to] && all[k].flow) {
      long long current = dfs(all[k].to, min(cap, all[k].flow));
      if (current == 0) continue;
      been[node] = 0;
      all[k].flow -= current;
      all[k ^ 1].flow += current;
      return current;
    }
  }
  return 0;
}
long long max_flow() {
  long long total_flow = 0;
  while (1) {
    bfs();
    if (bfs_dist[sink] == inf) break;
    memset(been, 0, sizeof(been));
    while (1) {
      long long current = dfs(source, inf);
      if (current == 0) break;
      total_flow += current;
    }
  }
  return total_flow;
}
int hsh(int node, int val) { return node * 207 + val + 104; }
struct f {
  int a, b, c;
};
f func[nmax];
int le[nmax], ri[nmax];
int eval(int which, int x) {
  return x * x * func[which].a + x * func[which].b + func[which].c;
}
int main() {
  scanf("%i%i", &n, &m);
  source = 0;
  sink = hsh(51, -101);
  for (int i = 1; i <= n; i++)
    scanf("%i%i%i", &func[i].a, &func[i].b, &func[i].c);
  for (int i = 1; i <= n; i++) scanf("%i%i", &le[i], &ri[i]);
  for (int i = 1; i <= n; i++) {
    add_edge(source, hsh(i, le[i]), inf);
    for (int j = le[i]; j <= ri[i]; j++)
      add_edge(hsh(i, j), hsh(i, j + 1), MX - eval(i, j));
    add_edge(hsh(i, ri[i] + 1), sink, inf);
  }
  int u, v, d;
  for (int i = 1; i <= m; i++) {
    scanf("%i%i%i", &u, &v, &d);
    for (int xu = le[u]; xu <= ri[u]; xu++) {
      int xv = xu - d;
      if (le[v] <= xv && xv <= ri[v] + 1) add_edge(hsh(u, xu), hsh(v, xv), inf);
    }
  }
  printf("%lld\n", MX * n - max_flow());
  return 0;
}