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 = 987654321;
const long long int INF = 123456789987654321;
FILE *fin, *fout;
int V, edge_cnt, source, sink;
vector<int> last_edge, nxt_edge, oppo, capa;
void init() {
  edge_cnt = 0, last_edge.clear(), nxt_edge.clear(), oppo.clear(), capa.clear(),
  last_edge = vector<int>(V, -1);
}
void add(int u, int v, int capa_) {
  nxt_edge.push_back(last_edge[u]), last_edge[u] = edge_cnt, oppo.push_back(v),
                                    capa.push_back(capa_), edge_cnt++;
}
vector<int> dist;
bool bfs() {
  dist.clear(), dist = vector<int>(V, -1);
  queue<int> q;
  q.push(source), dist[source] = 0;
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (int i = last_edge[now]; i != -1; i = nxt_edge[i]) {
      int nxt = oppo[i];
      if (capa[i] && dist[nxt] == -1) {
        dist[nxt] = dist[now] + 1;
        q.push(nxt);
      }
    }
  }
  if (dist[sink] == -1)
    return false;
  else
    return true;
}
vector<int> last_edge2;
int dfs(int now, int flow) {
  if (now == sink) return flow;
  for (int i = last_edge2[now]; i != -1; i = nxt_edge[i]) {
    last_edge2[now] = i;
    int nxt = oppo[i];
    if (capa[i] && dist[nxt] == dist[now] + 1) {
      if (int tmp = dfs(nxt, min(flow, capa[i]))) {
        capa[i] -= tmp;
        capa[i ^ 1] += tmp;
        return tmp;
      }
    }
  }
  return 0;
}
int dinic() {
  int total_flow = 0;
  while (bfs()) {
    last_edge2 = last_edge;
    while (int tmp = dfs(source, inf)) total_flow += tmp;
  }
  return total_flow;
}
int N, M;
struct Func {
  int a, b, c;
};
vector<Func> F;
vector<pair<int, int> > interval;
struct Restr {
  int u, v, d;
};
vector<Restr> restr;
int f(int idx, int x) { return F[idx].a * x * x + F[idx].b * x + F[idx].c; }
const int too_big = 10000000;
void get_network() {
  V = 220 * N + 2, source = V - 2, sink = V - 1;
  init();
  for (int i = 0; i < N; i++) {
    int l = interval[i].first, r = interval[i].second;
    add(source, 220 * i + (l - 1 + 110), inf);
    add(220 * i + (l - 1 + 110), source, 0);
    for (int j = 0; j < 219; j++) {
      add(220 * i + j, 220 * i + j + 1, too_big - f(i, j + 1 - 110));
      add(220 * i + j + 1, 220 * i + j, 0);
    }
    add(220 * i + (r + 110), sink, inf);
    add(sink, 220 * i + (r + 110), 0);
  }
  for (int i = 0; i < M; i++) {
    int u = restr[i].u, v = restr[i].v;
    for (int j = 0; j < 220; j++) {
      if (j - restr[i].d >= 0 && j - restr[i].d < 220) {
        add(220 * u + j, 220 * v + j - restr[i].d, inf);
        add(220 * v + j - restr[i].d, 220 * u + j, 0);
      }
    }
  }
  printf("%d", too_big * N - dinic());
}
int main() {
  scanf("%d %d", &N, &M);
  F.resize(N);
  for (int i = 0; i < N; i++) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    F[i] = {a, b, c};
  }
  interval.resize(N);
  for (int i = 0; i < N; i++) {
    scanf("%d %d", &interval[i].first, &interval[i].second);
  }
  restr.resize(M);
  for (int i = 0; i < M; i++) {
    int u, v, d;
    scanf("%d %d %d", &u, &v, &d);
    restr[i] = {--u, --v, d};
  }
  get_network();
}