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;
namespace MF {
int const N = 10000, M = 1000000;
int n, m, source, target;
int link[N], curr[N], dist[N];
int next[M], tail[M], cap[M], flow[M];
inline void add(int u, int v, int c) {
  m++;
  next[m] = link[u];
  link[u] = m;
  tail[m] = v;
  cap[m] = c;
  flow[m] = 0;
  m++;
  next[m] = link[v];
  link[v] = m;
  tail[m] = u;
  cap[m] = 0;
  flow[m] = 0;
}
inline bool bfs() {
  vector<int> queue;
  for (int i = 1; i <= n; ++i) {
    curr[i] = link[i];
    dist[i] = -1;
  }
  queue.push_back(target);
  dist[target] = 0;
  for (int i = 0; i < (int)queue.size(); ++i) {
    int u = queue[i];
    for (int e = link[u]; e > 0; e = next[e]) {
      int v = tail[e];
      if (flow[e ^ 1] < cap[e ^ 1] && dist[v] == -1) {
        dist[v] = dist[u] + 1;
        queue.push_back(v);
      }
    }
  }
  return dist[source] >= 0;
}
inline int dfs(int u, int ans) {
  if (u == target) {
    return ans;
  }
  int delta = ans;
  for (int& e = curr[u]; e > 0; e = next[e]) {
    int v = tail[e];
    if (flow[e] < cap[e] && dist[u] == dist[v] + 1) {
      int num = dfs(v, min(delta, cap[e] - flow[e]));
      flow[e] += num;
      flow[e ^ 1] -= num;
      delta -= num;
      if (delta == 0) {
        break;
      }
    }
  }
  return ans - delta;
}
inline int get() {
  int ans = 0, flow = 0;
  while (bfs()) {
    while (flow = dfs(source, INT_MAX)) {
      ans += flow;
    }
  }
  return ans;
}
}  // namespace MF
int const INF = 500000;
int const N = 100;
int n, m;
int a[N], b[N], c[N], l[N], r[N];
map<int, int> idx[N];
inline void build() {
  int num = 0;
  MF::source = ++num;
  MF::target = ++num;
  for (int i = 1; i <= n; ++i) {
    for (int j = l[i] - 1; j <= r[i]; ++j) {
      idx[i][j] = ++num;
    }
  }
  MF::n = num;
  MF::m = 1;
  for (int i = 1; i <= n; ++i) {
    MF::add(MF::source, idx[i][l[i] - 1], INT_MAX / 2);
    for (int j = l[i]; j <= r[i]; ++j) {
      int num = a[i] * j * j + b[i] * j + c[i];
      MF::add(idx[i][j - 1], idx[i][j], INF - num);
    }
    MF::add(idx[i][r[i]], MF::target, INT_MAX / 2);
  }
}
inline void limit(int u, int v, int d) {
  for (int i = l[u] - 1; i <= r[u]; ++i) {
    if (idx[v].count(i - d) > 0) {
      MF::add(idx[u][i], idx[v][i - d], INT_MAX / 2);
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d%d", a + i, b + i, c + i);
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", l + i, r + i);
  }
  build();
  for (int i = 1; i <= m; ++i) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    limit(u, v, d);
  }
  printf("%d\n", INF * n - MF::get());
  return 0;
}