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 = 200010, C = 500000, inf = 1 << 30;
int n, m, cnt = 1, cnt1, G[N];
struct iedge {
  int nx, t, f;
} E[N << 1];
int S, T;
int a[N], b[N], c[N], l[N], r[N];
int p[55][210];
inline int f(int x, int y) { return a[x] * y * y + b[x] * y + c[x]; }
inline void addedge(int x, int y, int f) {
  E[++cnt].t = y;
  E[cnt].nx = G[x];
  E[cnt].f = f;
  G[x] = cnt;
  E[++cnt].t = x;
  E[cnt].nx = G[y];
  E[cnt].f = 0;
  G[y] = cnt;
}
int dis[N];
queue<int> Q;
inline bool bfs() {
  while (!Q.empty()) Q.pop();
  memset(dis, -1, sizeof(dis));
  dis[S] = 0;
  Q.push(S);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = G[x]; i; i = E[i].nx)
      if (E[i].f && dis[E[i].t] == -1) {
        dis[E[i].t] = dis[x] + 1;
        Q.push(E[i].t);
        if (E[i].t == T) return true;
      }
  }
  return false;
}
int cur[N];
int dfs(int x, int f) {
  if (x == T || !f) return f;
  int used = 0, w;
  for (int i = G[x]; i; i = E[i].nx)
    if (E[i].f && dis[E[i].t] == dis[x] + 1) {
      w = dfs(E[i].t, min(E[i].f, f - used));
      E[i].f -= w;
      E[i ^ 1].f += w;
      if ((used += w) == f) return f;
    }
  return used;
}
int main() {
  scanf("%d%d", &n, &m);
  T = n * 210;
  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++) p[i][j + 100] = ++cnt1;
    addedge(S, p[i][l[i] + 100], C - f(i, l[i]));
    for (int j = l[i]; j < r[i]; j++)
      addedge(p[i][j + 100], p[i][j + 101], C - f(i, j + 1));
    addedge(p[i][r[i] + 100], T, inf);
  }
  for (int i = 1, u, v, d; i <= m; i++) {
    scanf("%d%d%d", &u, &v, &d);
    for (int j = l[v]; j <= r[v]; j++) {
      if (j + d > r[u]) continue;
      if (j + d >= l[u])
        addedge(p[u][j + d + 100], p[v][j + 100], inf);
      else
        addedge(S, p[v][j + 100], inf);
    }
  }
  int ans = 0;
  while (bfs()) memcpy(cur, G, sizeof(cur)), ans += dfs(S, inf);
  printf("%d\n", n * C - ans);
  return 0;
}