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 maxI = 51;
const int maxK = 210;
const int Num = 100;
const int maxN = maxI * maxK;
const int maxM = maxN * 10;
const int inf = 1000000000;
const long long INF = 1e18;
class Edge {
 public:
  int v;
  long long flow;
};
int n, m, S, T;
int A[maxI], B[maxI], C[maxI];
long long Key[maxK];
int Id[maxI][maxK];
int ecnt = -1, Hd[maxN], Nt[maxM];
Edge E[maxM];
int cur[maxN], Qu[maxN], label[maxN];
void Add_Edge(int u, int v, long long flow);
bool Bfs();
long long dfs(int u, long long flow);
int main() {
  scanf("%d%d", &n, &m);
  memset(Hd, -1, sizeof(Hd));
  for (int i = 1; i <= n; i++) scanf("%d%d%d", &A[i], &B[i], &C[i]);
  int idcnt = 0;
  for (int i = 1; i <= n; i++)
    for (int j = -100; j < 100; j++) Id[i][j + Num] = ++idcnt;
  S = ++idcnt;
  T = ++idcnt;
  for (int i = 1; i <= n; i++) Id[i][-101 + Num] = S, Id[i][100 + Num] = T;
  long long Ans = 0;
  for (int i = 1; i <= n; i++) {
    int l, r;
    long long sum = 0;
    scanf("%d%d", &l, &r);
    for (int j = -100; j <= 100; j++)
      if (j < l || j > r)
        Key[j + Num] = 0;
      else
        Key[j + Num] = j * j * A[i] + B[i] * j + C[i] + inf,
                sum += Key[j + Num];
    Ans += sum;
    for (int j = -100; j <= 100; j++)
      Add_Edge(Id[i][j + Num - 1], Id[i][j + Num], sum - Key[j + Num]);
  }
  for (int i = 1; i <= m; i++) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    for (int j = d - 100; j <= 100; j++)
      Add_Edge(Id[u][j + Num], Id[v][j - d + Num], INF);
  }
  while (Bfs()) {
    memcpy(cur, Hd, sizeof(cur));
    while (long long di = dfs(S, INF)) Ans -= di;
  }
  printf("%I64d\n", Ans - 1ll * n * inf);
  return 0;
}
void Add_Edge(int u, int v, long long flow) {
  Nt[++ecnt] = Hd[u];
  Hd[u] = ecnt;
  E[ecnt] = ((Edge){v, flow});
  Nt[++ecnt] = Hd[v];
  Hd[v] = ecnt;
  E[ecnt] = ((Edge){u, 0});
  return;
}
bool Bfs() {
  memset(label, -1, sizeof(label));
  int ql = 1, qr = 1;
  Qu[1] = S;
  label[S] = 1;
  while (ql <= qr)
    for (int u = Qu[ql++], i = Hd[u]; i != -1; i = Nt[i])
      if (E[i].flow && label[E[i].v] == -1)
        label[Qu[++qr] = E[i].v] = label[u] + 1;
  return label[T] != -1;
}
long long dfs(int u, long long flow) {
  if (u == T) return flow;
  for (int &i = cur[u]; i != -1; i = Nt[i])
    if (E[i].flow && label[E[i].v] == label[u] + 1)
      if (long long di = dfs(E[i].v, min(flow, E[i].flow)))
        return E[i].flow -= di, E[i ^ 1].flow += di, di;
  return 0;
}