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 INF = 1e9;
const int MAXN = 150005;
const int MAXM = 1900000;
int N, edges;
int last[MAXN], preve[MAXM], head[MAXM];
int Cap[MAXM], Flow[MAXM];
int dist[MAXN];
int nextEdge[MAXN];
int edgeno[MAXN];
queue<int> Q;
void init() {
  edges = 0;
  memset(last, -1, sizeof last);
}
inline void add_edge(int u, int v, int cap, int flow = 0) {
  head[edges] = v;
  preve[edges] = last[u];
  Cap[edges] = cap;
  Flow[edges] = flow;
  last[u] = edges++;
  head[edges] = u;
  preve[edges] = last[v];
  Cap[edges] = 0;
  Flow[edges] = 0;
  last[v] = edges++;
}
inline bool dinicBfs(int S, int E, int N) {
  int from = S, to, cap, flow;
  memset(dist, 0, sizeof(int) * N);
  dist[from] = 1;
  while (!Q.empty()) Q.pop();
  Q.push(from);
  while (!Q.empty()) {
    from = Q.front();
    Q.pop();
    for (int e = last[from]; e >= 0; e = preve[e]) {
      to = head[e];
      cap = Cap[e];
      flow = Flow[e];
      if (!dist[to] && cap > flow) {
        dist[to] = dist[from] + 1;
        Q.push(to);
        if (to == E) return true;
      }
    }
  }
  return (dist[E] != 0);
}
inline int dfs(int from, int minEdge, int E) {
  if (!minEdge) return 0;
  if (from == E) return minEdge;
  int to, e, cap, flow, ret;
  for (; nextEdge[from] >= 0; nextEdge[from] = preve[e]) {
    e = nextEdge[from];
    to = head[e];
    cap = Cap[e];
    flow = Flow[e];
    if (dist[to] != dist[from] + 1) continue;
    ret = dfs(to, min(minEdge, cap - flow), E);
    if (ret) {
      Flow[e] += ret;
      Flow[e ^ 1] -= ret;
      return ret;
    }
  }
  return 0;
}
int dinicUpdate(int S, int E) {
  int flow = 0;
  while (int minEdge = dfs(S, INF, E)) {
    if (minEdge == 0) break;
    flow += minEdge;
  }
  return flow;
}
int maxFlow(int S, int E, int N) {
  int totFlow = 0;
  while (dinicBfs(S, E, N)) {
    for (int i = 0; i <= N; i++) nextEdge[i] = last[i];
    totFlow += dinicUpdate(S, E);
  }
  return totFlow;
}
int a[70], b[70], c[70], L[70], R[70];
map<int, int> name[70];
int get(int a, int b, int c, int j) {
  int sum = a * (j * j);
  sum += b * j;
  sum += c;
  return sum;
}
int main() {
  int i, j, k, l, m, n, maxi = INT_MIN, cnt = 0;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
  }
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &L[i], &R[i]);
    sum += (R[i] - L[i] + 2);
  }
  maxi = (1e7) + 5;
  init();
  int sink = sum + 1, source = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = L[i] - 1; j <= R[i]; j++) {
      name[i][j] = ++cnt;
      if (j < L[i]) {
        add_edge(source, cnt, INF, 0);
      } else {
        edgeno[cnt] = edges;
        add_edge(cnt - 1, cnt, maxi - get(a[i], b[i], c[i], j));
      }
    }
    add_edge(cnt, sink, INF);
  }
  assert(cnt < sink);
  for (int i = 1; i <= m; i++) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    for (int j = L[u]; j <= R[u]; j++) {
      if ((j - d) > R[v]) {
        int c = name[u][j - 1];
        int g = name[v][R[v]];
        add_edge(c, g, INF, 0);
        break;
      }
      if ((j - d) < L[v]) continue;
      int c = name[u][j - 1];
      int e = name[v][j - d - 1];
      add_edge(c, e, INF, 0);
    }
  }
  int ans = maxFlow(source, sink, sink + 2);
  ans = maxi * n - ans;
  cout << ans << endl;
}