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 double PI = acos(-1.0);
const long long INF = 1e9;
const long long LINF = 1e18;
namespace MaxFlow {
const int MAX = 1 << 20;
struct edge {
  int from;
  int to;
  long long cap;
  long long flow;
};
int n;
int source;
int sink;
int d[MAX];
int ptr[MAX];
vector<edge> e;
vector<int> g[MAX];
void add(int u, int v, long long w) {
  g[u].push_back((int)((e).size()));
  e.push_back({u, v, w, 0});
  g[v].push_back((int)((e).size()));
  e.push_back({v, u, 0, 0});
}
bool bfs() {
  queue<int> q;
  q.push(source);
  for (int i = (0); i < (n); ++i) d[i] = -1;
  d[source] = 0;
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (auto i : g[v]) {
      int to = e[i].to;
      if (d[to] != -1) continue;
      if (e[i].flow == e[i].cap) continue;
      q.push(to);
      d[to] = d[v] + 1;
    }
  }
  return d[sink] != -1;
}
long long dfs(int v, long long flow) {
  if (!flow) return 0;
  if (v == sink) return flow;
  while (1) {
    if (ptr[v] == (int)((g[v]).size())) break;
    int id = g[v][ptr[v]];
    int to = e[id].to;
    if (d[to] != d[v] + 1) {
      ++ptr[v];
      continue;
    }
    long long push = dfs(to, min(flow, e[id].cap - e[id].flow));
    if (push == 0) {
      ++ptr[v];
      continue;
    }
    e[id].flow += push;
    e[id ^ 1].flow -= push;
    return push;
  }
  return 0;
}
long long dinic() {
  cerr << "Don't forget about constant: " << MAX << endl;
  cerr << "Check n: " << n << endl;
  long long flow = 0;
  while (1) {
    if (!bfs()) break;
    for (int i = (0); i < (n); ++i) ptr[i] = 0;
    while (long long push = dfs(source, LINF)) flow += push;
  }
  return flow;
}
};  // namespace MaxFlow
const int MAX = 100;
int n, m;
int a[MAX], b[MAX], c[MAX], l[MAX], r[MAX];
int node(int i, int pos) { return 205 * (i + 1) + pos - (l[i] - 1); }
int F(int i, int j) { return a[i] * j * j + b[i] * j + c[i]; }
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  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];
  MaxFlow::source = 0;
  MaxFlow::sink = 50 * 210;
  MaxFlow::n = 50 * 210 + 1;
  for (int i = (0); i < (n); ++i) {
    MaxFlow::add(MaxFlow::source, node(i, l[i] - 1), LINF);
    MaxFlow::add(node(i, r[i]), MaxFlow::sink, LINF);
    for (int j = (l[i] - 1); j < (r[i]); ++j)
      MaxFlow::add(node(i, j), node(i, j + 1), INF - F(i, j + 1));
  }
  int u, v, w;
  for (int i = (0); i < (m); ++i) {
    cin >> u >> v >> w;
    --u;
    --v;
    for (int j = (l[u] - 1); j < (r[u] + 1); ++j) {
      int x = j - w;
      if (l[v] - 1 <= x && r[v] >= x)
        MaxFlow::add(node(u, j), node(v, x), LINF);
    }
  }
  long long Flow = MaxFlow::dinic();
  long long ans = n * INF;
  ans -= Flow;
  cout << ans << endl;
  cerr << "Time elapsed: " << clock() / (double)CLOCKS_PER_SEC << endl;
  return 0;
}