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 = 0x3f3f3f3f;
const int MAX = 1e6;
namespace flow {
const int N = 3e4, INF = 0x3f3f3f3f;
int SOURCE = N - 1, SINK = N - 2;
struct Edge {
  int u = 0, r = 0, c = 0, f = 0;
  Edge(int aa, int bb, int cc, int dd) : u(aa), r(bb), c(cc), f(dd) {}
};
vector<Edge> g[N];
void add(int a, int b, int c) {
  g[a].emplace_back(b, (int)g[b].size(), c, 0);
  g[b].emplace_back(a, (int)g[a].size() - 1, 0, 0);
}
int lim, dist[N], p[N];
bool bfs() {
  memset(dist, 0x3f, sizeof dist);
  memset(p, 0x00, sizeof p);
  queue<int> q;
  q.push(SOURCE);
  dist[SOURCE] = 0;
  while (q.size()) {
    int v = q.front();
    q.pop();
    for (auto &e : g[v]) {
      if (dist[e.u] > dist[v] + 1 && e.c - e.f >= lim) {
        q.push(e.u);
        dist[e.u] = dist[v] + 1;
      }
    }
  }
  return dist[SINK] < INF;
}
int dfs(int v, int min_c) {
  if (v == SINK) {
    return min_c;
  }
  for (; p[v] < (int)g[v].size(); ++p[v]) {
    auto &e = g[v][p[v]];
    if (dist[e.u] > dist[v] && e.c - e.f >= lim) {
      int d = dfs(e.u, min(min_c, e.c - e.f));
      if (d > 0) {
        e.f += d;
        g[e.u][e.r].f -= d;
        return d;
      }
    }
  }
  return 0;
}
long long find_flow() {
  long long answer = 0;
  lim = 1;
  for (lim = 1 << 30; lim > 0; lim /= 2) {
    while (bfs()) {
      int d = dfs(SOURCE, INF);
      while (d > 0) {
        answer += d;
        d = dfs(SOURCE, INF);
      }
    }
  }
  return answer;
}
}  // namespace flow
int id(int i, int x) {
  x += 100;
  assert(0 <= i && i < 50);
  assert(0 <= x && x < 202);
  return 205 * i + x;
}
const int N = 100;
int a[N], b[N], c[N];
int l[N], r[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  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];
    flow::add(flow::SOURCE, id(i, -100), INF);
    for (int x = -100; x < l[i]; ++x) {
      flow::add(id(i, x), id(i, x + 1), INF);
    }
    for (int x = l[i]; x <= r[i]; ++x) {
      int f = a[i] * x * x + b[i] * x + c[i];
      flow::add(id(i, x), id(i, x + 1), MAX - f);
    }
    for (int x = r[i] + 1; x <= 100; ++x) {
      flow::add(id(i, x), id(i, x + 1), INF);
    }
    flow::add(id(i, 101), flow::SINK, INF);
  }
  while (m--) {
    int i, j, d;
    cin >> i >> j >> d;
    --i;
    --j;
    for (int xi = -100; xi <= 100; ++xi) {
      int xj = max(min(xi - d, 101), -100);
      flow::add(id(i, xi), id(j, xj), INF);
    }
  }
  long long min_cut = flow::find_flow();
  long long answer = (long long)MAX * n - min_cut;
  cout << answer << '\n';
  return 0;
}