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 MCMF {
int tot = 1;
int head[200010], dep[200010];
struct edge {
  int to, next;
  long long len;
} dat[200010];
void add_edge(int x, int y, long long z) {
  dat[++tot] = (edge){y, head[x], z};
  head[x] = tot;
}
void Add_edge(int x, int y, long long z) {
  add_edge(x, y, z);
  add_edge(y, x, 0);
}
int fuck[200010], S, T;
queue<int> q;
int bfs() {
  for (int i = S; i <= T; i++) dep[i] = 0, fuck[i] = head[i];
  dep[S] = 1;
  q.push(S);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i; i = dat[i].next) {
      int t = dat[i].to;
      if (!dat[i].len || dep[t]) continue;
      dep[t] = dep[x] + 1;
      q.push(t);
    }
  }
  return dep[T];
}
long long dfs(int x, long long limit) {
  if (x == T) return limit;
  long long sum = 0;
  for (int i = fuck[x]; i; i = dat[i].next) {
    fuck[x] = i;
    int t = dat[i].to;
    if (!dat[i].len || dep[t] != dep[x] + 1) continue;
    long long hh = dfs(t, min(limit, dat[i].len));
    dat[i].len -= hh;
    dat[i ^ 1].len += hh;
    sum += hh;
    limit -= hh;
    if (!limit) return sum;
  }
  return sum;
}
long long dinic() {
  long long sum = 0;
  while (bfs()) {
    sum += dfs(S, 1e18);
  }
  return sum;
}
void init(int _S, int _T) {
  S = _S, T = _T;
  for (int i = S; i <= T; i++) head[i] = 0;
  tot = 1;
}
}  // namespace MCMF
using MCMF::Add_edge;
using MCMF::dinic;
long long f(long long x, long long a, long long b, long long c) {
  return x * x * a + x * b + c;
}
const int C = 1000000;
int n, m, S, T, a[90100], b[90100], c[90100], pos[90100], l[90100], r[90100],
    id[60][300], cnt;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) scanf("%d%d%d", &a[i], &b[i], &c[i]);
  S = 0, T = 100000;
  MCMF::init(S, T);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    for (int j = l[i]; j <= r[i] + 1; j++) {
      id[i][j + 100] = ++cnt;
      if (j == l[i])
        Add_edge(S, id[i][j + 100], 1e18);
      else
        Add_edge(id[i][j - 1 + 100], id[i][j + 100],
                 C - f(j - 1, a[i], b[i], c[i]));
    }
    Add_edge(id[i][r[i] + 1 + 100], T, 1e18);
  }
  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] + 1 && j - d >= l[v])
        Add_edge(id[u][j + 100], id[v][j + 100 - d], 1e18);
    }
  }
  long long fuck = dinic();
  cout << -fuck + 1ll * n * C;
}