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;
struct FlowNetwork {
  const int INF = INT_MAX / 2;
  struct Edge {
    int u, v;
    int capacity;
    int rev;
  };
  int n;
  vector<Edge> edges;
  vector<vector<int>> adj;
  vector<int> rev;
  FlowNetwork(int _n) {
    n = _n;
    edges.clear();
    rev.clear();
    adj.resize(n);
    for (int i = 0; i < n; i++) adj[i].clear();
  }
  void addEdge(int u, int v, int capacity) {
    u--;
    v--;
    edges.push_back(Edge{u, v, capacity, 0});
    edges.push_back(Edge{v, u, 0, 0});
    adj[u].push_back((int)edges.size() - 2);
    adj[v].push_back((int)edges.size() - 1);
    edges.end()[-2].rev = (int)edges.size() - 1;
    edges.end()[-1].rev = (int)edges.size() - 2;
  }
  vector<int> level;
  vector<int> first;
  bool bfs() {
    level.resize(n);
    for (int i = 0; i < n; i++) level[i] = -1;
    queue<int> q;
    level[0] = 0;
    q.push(0);
    while (q.empty() == false) {
      int u = q.front();
      q.pop();
      for (int i : adj[u])
        if (edges[i].capacity > 0) {
          if (level[edges[i].v] == -1) {
            level[edges[i].v] = level[u] + 1;
            q.push(edges[i].v);
          }
        }
    }
    return level[n - 1] != -1;
  }
  int dfs(int u, int pushed) {
    if (u == n - 1) return pushed;
    while (first[u] < (int)adj[u].size()) {
      int i = adj[u][first[u]];
      first[u]++;
      if (level[u] + 1 == level[edges[i].v] && edges[i].capacity > 0) {
        int push = dfs(edges[i].v, min(pushed, edges[i].capacity));
        if (push == 0) continue;
        edges[i].capacity -= push;
        edges[edges[i].rev].capacity += push;
        return push;
      }
    }
    return 0;
  }
  int maxFlow() {
    int answer = 0;
    first.resize(n);
    while (bfs() == true) {
      for (int i = 0; i < n; i++) first[i] = 0;
      while (true) {
        int push = dfs(0, INF);
        if (push == 0) break;
        answer += push;
      }
    }
    return answer;
  }
};
const int N_MAX = 50;
const int M_MAX = 100;
const int F_MAX = 300000;
int n, m;
struct QuadFun {
  int a, b, c;
  int eval(int x) { return a * x * x + b * x + c; }
};
QuadFun f[N_MAX + 2];
int l[N_MAX + 2], r[N_MAX + 2];
int firstNode[N_MAX + 2];
int codif(int i, int x) { return firstNode[i] + (x - (l[i] - 1)); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> f[i].a >> f[i].b >> f[i].c;
  for (int i = 1; i <= n; i++) cin >> l[i] >> r[i];
  int netSize = 2;
  for (int i = 1; i <= n; i++) netSize += r[i] - (l[i] - 1) + 1;
  FlowNetwork fnet(netSize);
  int currNode = 1;
  for (int i = 1; i <= n; i++) {
    currNode++;
    fnet.addEdge(1, currNode, fnet.INF);
    firstNode[i] = currNode;
    for (int x = l[i]; x <= r[i]; x++) {
      currNode++;
      fnet.addEdge(currNode - 1, currNode, F_MAX - f[i].eval(x));
    }
    fnet.addEdge(currNode, netSize, fnet.INF);
  }
  for (int i = 1; i <= m; i++) {
    int u, v, d;
    cin >> u >> v >> d;
    for (int x = l[u] - 1; x <= r[u]; x++)
      if (l[v] - 1 <= x - d && x - d <= r[v])
        fnet.addEdge(codif(u, x), codif(v, x - d), fnet.INF);
  }
  cout << n * F_MAX - fnet.maxFlow() << "\n";
  return 0;
}