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 Edge {
  int u, v, cap, flow;
};
const int MAXN = 5000005;
int N, d[MAXN], ptr[MAXN], q[MAXN];
vector<Edge> edge;
vector<int> g[MAXN];
void add_edge(int u, int v, int cap) {
  Edge e1 = {u, v, cap, 0};
  Edge e2 = {v, u, 0, 0};
  g[u].push_back((int)edge.size());
  edge.push_back(e1);
  g[v].push_back((int)edge.size());
  edge.push_back(e2);
}
bool bfs(const int source, const int sink) {
  memset(d, -1, N * sizeof(d[0]));
  d[source] = 0;
  int qh = 0, qt = 0;
  q[qt++] = source;
  while (qh != qt && d[sink] == -1) {
    int u = q[qh++];
    for (size_t i = 0; i < g[u].size(); ++i) {
      int id = g[u][i];
      int v = edge[id].v;
      if (d[v] == -1 && edge[id].flow < edge[id].cap) {
        d[v] = d[u] + 1;
        q[qt++] = v;
      }
    }
  }
  return d[sink] != -1;
}
int dfs(const int u, const int t, const int flow) {
  if (flow == 0) return 0;
  if (u == t) return flow;
  for (int &i = ptr[u]; i < g[u].size(); ++i) {
    int id = g[u][i];
    int v = edge[id].v;
    if (d[v] != d[u] + 1) continue;
    int pushed = dfs(v, t, min(flow, edge[id].cap - edge[id].flow));
    if (pushed) {
      edge[id].flow += pushed;
      edge[id ^ 1].flow -= pushed;
      return pushed;
    }
  }
  return 0;
}
int dinic(int source, int sink) {
  int flow = 0;
  while (bfs(source, sink)) {
    memset((ptr), (0), sizeof(ptr));
    while (int push = dfs(source, sink, 1011111111)) flow += push;
  }
  return flow;
}
const int MAX_N = 50 + 1, MAX_M = 100 + 1;
const int BIG = int(3e5);
const int MX = 102;
int a[MAX_N], b[MAX_N], c[MAX_N];
int l[MAX_M], r[MAX_M];
vector<int> id[MAX_N];
int F(int i, int x) { return (a[i] * x + b[i]) * x + c[i]; }
int get(int u, int i) {
  if (i < -MX || i > MX) return -1;
  return id[u][i + MX];
}
void addEdge(int u, int i, int v, int j) {
  int x = get(u, i), y = get(v, j);
  if (x == -1 || y == -1) return;
  add_edge(x, y, 1011111111);
}
void constraint(int u, int v, int d) {
  for (int i = -MX; i <= MX; ++i) {
    addEdge(v, i, u, i + d);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int _n(n), i(0); i < _n; i++) cin >> a[i] >> b[i] >> c[i];
  for (int _n(n), i(0); i < _n; i++) cin >> l[i] >> r[i];
  int S = 0, T = 1;
  int V = 2;
  int total = 2 * MX + 2;
  for (int _n(n), me(0); me < _n; me++) {
    vector<int> &v = id[me];
    for (int _n(total), i(0); i < _n; i++) v.push_back(V++);
    for (int i = -MX; i <= MX; ++i) {
      if (i >= l[me] && i <= r[me])
        add_edge(v[i + MX], v[i + MX + 1], BIG - F(me, i));
      else
        add_edge(v[i + MX], v[i + MX + 1], 1011111111);
    }
    add_edge(S, v[0], 1011111111);
    add_edge(v.back(), T, 1011111111);
  }
  while (m--) {
    int u, v, d;
    cin >> u >> v >> d;
    --u, --v;
    constraint(v, u, -d);
  }
  N = V;
  int ans = BIG * n - dinic(S, T);
  cout << ans << endl;
  return 0;
}