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 = 1 << 30;
const int N = 20100;
const int M = 800100;
struct Edge {
  int from, to, cap;
  Edge(int from = 0, int to = 0, int cap = 0) : from(from), to(to), cap(cap) {}
} egs[M];
int G[N], ne[M], xb, cr[N];
void add(int from, int to, int cap) {
  ne[xb] = G[from];
  egs[xb] = Edge(from, to, cap);
  G[from] = xb++;
  ne[xb] = G[to];
  egs[xb] = Edge(to, from, 0);
  G[to] = xb++;
}
int t_, w_, q[N], d[N];
int S, T;
bool bfs() {
  for (memset(d, -1, sizeof d), t_ = -1, q[w_ = d[T] = 0] = T; t_ < w_;)
    for (int x = q[++t_], i = G[x]; ~i; i = ne[i])
      if (!~d[egs[i].to] && egs[i ^ 1].cap) {
        d[q[++w_] = egs[i].to] = d[x] + 1;
        if (egs[i].to == S) return 1;
      }
  return ~d[S];
}
int dfs(int x, int a) {
  if (x == T || !a) return a;
  int fw = 0, f;
  for (int& i = cr[x]; ~i; i = ne[i])
    if (d[x] - 1 == d[egs[i].to] &&
        (f = dfs(egs[i].to, min(a, egs[i].cap))) > 0) {
      egs[i].cap -= f;
      egs[i ^ 1].cap += f;
      fw += f;
      a -= f;
      if (!a) break;
    }
  return fw;
}
int mf() {
  int fw = 0;
  for (; bfs(); memcpy(cr, G, sizeof G), fw += dfs(S, inf))
    ;
  return fw;
}
int n, m, cnt;
int a[N], b[N], c[N], id[100][300], l[N], r[N], sum;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  S = 1;
  T = 2;
  cnt = 2;
  memset(G, -1, sizeof G);
  for (int i = 1; i <= n; ++i) cin >> a[i] >> b[i] >> c[i];
  for (int i = 1; i <= n; ++i) {
    cin >> l[i] >> r[i];
    int mx = INT_MIN;
    for (int j = l[i]; j <= r[i]; ++j)
      mx = max(mx, a[i] * j * j + b[i] * j + c[i]);
    sum += mx;
    for (int j = 0; j <= r[i] - l[i] + 1; ++j) id[i][j] = ++cnt;
    add(S, id[i][0], inf);
    add(id[i][r[i] - l[i] + 1], T, inf);
    for (int j = 0; j <= r[i] - l[i]; ++j)
      add(id[i][j], id[i][j + 1],
          mx - ((a[i] * (j + l[i]) + b[i]) * (j + l[i]) + c[i]));
  }
  for (int i = 1; i <= m; ++i) {
    int x, y, z;
    cin >> x >> y >> z;
    for (int j = l[x]; j <= r[x]; ++j)
      if (j - z >= l[y])
        add(id[x][j - l[x]], id[y][min(r[y] + 1, j - z) - l[y]], inf);
  }
  cout << (sum -= mf()) << '\n';
  return 0;
}