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 {
  long long a, b;
  long long cap, flow;
  edge(long long az, long long bz, long long capz, long long flowz) {
    a = az;
    b = bz;
    cap = capz;
    flow = flowz;
  }
};
vector<edge> ed;
vector<long long> g[(long long)3e5];
long long d[(long long)3e5];
long long pos[(long long)3e5];
void add(long long a, long long b, long long c) {
  g[a].push_back(ed.size());
  ed.push_back(edge(a, b, c, 0));
  g[b].push_back(ed.size());
  ed.push_back(edge(b, a, 0, 0));
}
long long INF = 1e18;
long long M = 1e12;
bool bfs(long long s, long long t, long long B) {
  queue<long long> q;
  for (long long i = s; i <= t; i++) d[i] = INF;
  d[s] = 0;
  q.push(s);
  while (!q.empty()) {
    long long v = q.front();
    q.pop();
    for (auto x : g[v]) {
      if (d[v] + 1 < d[ed[x].b] && ed[x].cap - ed[x].flow >= B) {
        d[ed[x].b] = d[v] + 1;
        q.push(ed[x].b);
      }
    }
  }
  return (d[t] < INF);
}
long long ans = 0;
long long cnt = 0;
long long dfs(long long v, long long t, long long cmin, long long B) {
  if (v == t) {
    ans += cmin;
    return cmin;
  }
  for (; pos[v] < (long long)g[v].size(); pos[v]++) {
    long long x = g[v][pos[v]];
    if (d[ed[x].a] + 1 == d[ed[x].b] && ed[x].cap - ed[x].flow >= B) {
      long long d = dfs(ed[x].b, t, min(cmin, ed[x].cap - ed[x].flow), B);
      if (d > 0) {
        ed[x].flow += d;
        ed[x ^ 1].flow -= d;
        return d;
      }
    }
  }
  return 0;
}
long long S, T;
const long long sz = 210;
long long a[100];
long long b[100];
long long c[100];
long long l[100];
long long r[100];
int main() {
  long long n, m;
  cin >> n >> m;
  S = 0;
  T = (n + 1) * sz;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i] >> b[i] >> c[i];
  }
  for (long long i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
    add(S, l[i] + i * sz - 1, INF);
    for (long long j = l[i] + i * sz; j <= r[i] + i * sz; j++) {
      add(j - 1, j,
          M - a[i] * (j - i * sz) * (j - i * sz) - b[i] * (j - i * sz) - c[i]);
    }
    add(r[i] + i * sz, T, INF);
  }
  for (long long i = 1; i <= m; i++) {
    long long u, v, e;
    cin >> u >> v >> e;
    for (long long j = l[v] - 1; j <= r[v]; j++) {
      if (j + e <= r[u] && j + e >= l[u] - 1)
        add(u * sz + j + e, v * sz + j, INF);
    }
  }
  for (long long i = 30; i >= 0; i--) {
    long long B = (1ll << i);
    while (bfs(S, T, B)) {
      for (long long j = 0; j <= T; j++) pos[j] = 0;
      while (dfs(S, T, INF, B)) {
      }
    }
  }
  cout << M * n - ans << endl;
}