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 LIM = 1000000;
int n, m, a[110], b[110], c[110], L[110], R[110], id[110][2100], tot;
namespace MaxFlow {
const int N = 201000, M = 2001000;
int head[N], cur[N], dep[N], cnt, S, T, res;
struct node {
  int to, next, val;
} edge[M];
void ae(int u, int v, int w) {
  edge[cnt].next = head[u], edge[cnt].to = v, edge[cnt].val = w,
  head[u] = cnt++;
  edge[cnt].next = head[v], edge[cnt].to = u, edge[cnt].val = 0,
  head[v] = cnt++;
}
queue<int> q;
inline bool bfs() {
  memset(dep, 0, sizeof(dep)), q.push(S), dep[S] = 1;
  while (!q.empty()) {
    register int x = q.front();
    q.pop();
    for (register int i = cur[x] = head[x]; i != -1; i = edge[i].next)
      if (edge[i].val && !dep[edge[i].to])
        dep[edge[i].to] = dep[x] + 1, q.push(edge[i].to);
  }
  return dep[T] > 0;
}
bool reach;
inline int dfs(int x, int flow) {
  if (x == T) {
    res += flow, reach = true;
    return flow;
  }
  int used = 0;
  for (register int &i = cur[x]; i != -1; i = edge[i].next) {
    if (!edge[i].val || dep[edge[i].to] != dep[x] + 1) continue;
    register int ff = dfs(edge[i].to, min(edge[i].val, flow - used));
    if (ff) {
      edge[i].val -= ff, edge[i ^ 1].val += ff, used += ff;
      if (used == flow) break;
    }
  }
  return used;
}
inline void Dinic() {
  while (bfs()) {
    reach = true;
    while (reach) reach = false, dfs(S, 0x3f3f3f3f);
  }
}
}  // namespace MaxFlow
using namespace MaxFlow;
int main() {
  scanf("%d%d", &n, &m), S = 1, T = 2, tot = 2, memset(head, -1, sizeof(head));
  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]);
    for (int j = L[i]; j <= R[i]; j++) id[i][(j) + L[i]] = ++tot;
    ae(S, id[i][(L[i]) + L[i]],
       LIM - (a[i] * (L[i]) * (L[i]) + b[i] * (L[i]) + c[i]));
    for (int j = L[i] + 1; j <= R[i]; j++)
      ae(id[i][(j - 1) + L[i]], id[i][(j) + L[i]],
         LIM - (a[i] * (j) * (j) + b[i] * (j) + c[i]));
    ae(id[i][(R[i]) + L[i]], T, 0x3f3f3f3f);
  }
  for (int i = 1, u, v, d; i <= m; i++) {
    scanf("%d%d%d", &u, &v, &d);
    if (L[v] <= L[u] - d - 1 && L[u] - d - 1 <= R[v])
      ae(S, id[v][(L[u] - d - 1) + L[v]], 0x3f3f3f3f);
    for (int j = L[u]; j <= R[u]; j++) {
      if (j - d < L[v]) continue;
      if (j - d <= R[v])
        ae(id[u][(j) + L[u]], id[v][(j - d) + L[v]], 0x3f3f3f3f);
      else
        ae(id[u][(j) + L[u]], T, 0x3f3f3f3f);
    }
  }
  Dinic();
  printf("%d\n", LIM * n - res);
  return 0;
}