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 N = 21000 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const int MAX = 2e5;
int head[N], level[N], tot, S, T;
int n, m, a[N], b[N], c[N], L[N], R[N];
struct node {
  int to, w, nx;
} edge[2000007];
void add(int u, int v, int w) {
  edge[tot].to = v;
  edge[tot].w = w;
  edge[tot].nx = head[u];
  head[u] = tot++;
  edge[tot].to = u;
  edge[tot].w = 0;
  edge[tot].nx = head[v];
  head[v] = tot++;
}
bool bfs() {
  memset(level, 0, sizeof(level));
  queue<int> que;
  level[S] = 1;
  que.push(S);
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    if (u == T) return true;
    for (int i = head[u]; ~i; i = edge[i].nx) {
      int v = edge[i].to, w = edge[i].w;
      if (level[v] || w <= 0) continue;
      level[v] = level[u] + 1;
      que.push(v);
    }
  }
  return false;
}
int dfs(int u, int p) {
  if (u == T) return p;
  int ret = 0;
  for (int i = head[u]; ~i; i = edge[i].nx) {
    int v = edge[i].to, w = edge[i].w;
    if (level[v] != level[u] + 1 || w <= 0) continue;
    int f = dfs(v, min(w, p - ret));
    ret += f;
    edge[i].w -= f;
    edge[i ^ 1].w += f;
    if (ret == p) break;
  }
  if (!ret) level[u] = 0;
  return ret;
}
int Dinic() {
  int ans = 0;
  while (bfs()) ans += dfs(S, inf);
  return ans;
}
void init() {
  memset(head, -1, sizeof(head));
  tot = 0;
}
inline int ID(int x, int y) { return x * 208 + y + 102; }
inline int getVal(int who, int x) {
  return a[who] * x * x + b[who] * x + c[who];
}
int main() {
  init();
  S = 21000, T = 0;
  cin >> n >> m;
  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];
    add(S, ID(i, L[i] - 1), inf);
    for (int j = L[i]; j <= R[i]; j++)
      add(ID(i, j - 1), ID(i, j), MAX - getVal(i, j));
    add(ID(i, R[i]), T, inf);
  }
  while (m--) {
    int u, v, d;
    cin >> u >> v >> d;
    for (int j = L[u]; j <= R[u]; j++)
      if (j - d >= L[v] && j - d - 1 <= R[v])
        add(ID(u, j - 1), ID(v, j - d - 1), inf);
  }
  printf("%d\n", n * MAX - Dinic());
  return 0;
}