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;
struct point {
  int a, b, c, l, r;
  inline int y(int x) { return x * x * a + x * b + c; }
} p[52];
struct edge {
  int next, other, va;
} e[100010];
int num[52][210], n, m, S, T, head[11111], pre[11111], vh[11111], di[11111],
    dis[11111], his[11111];
int inf = 100000000, tot, ans, Max, u, v, d;
inline void read(int& x) {
  char ch = getchar();
  int mk = 1;
  for (; (ch < '0' || ch > '9') && ch != '-'; ch = getchar())
    ;
  if (ch == '-') mk = -1, ch = getchar();
  for (x = 0; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - 48;
  x *= mk;
}
inline void add(int x, int y, int z) {
  e[++tot].next = head[x];
  head[x] = tot;
  e[tot].other = y;
  e[tot].va = z;
  e[++tot].next = head[y];
  head[y] = tot;
  e[tot].other = x;
  e[tot].va = 0;
}
void flow() {
  for (int i = S; i <= T; i++) di[i] = head[i];
  vh[0] = T;
  int u = S, aug = inf;
  while (dis[S] < T) {
    bool flag = 0;
    his[u] = aug;
    for (int i = di[u]; i; i = e[i].next) {
      int v = e[i].other;
      if (e[i].va > 0 && dis[u] == dis[v] + 1) {
        pre[v] = di[u] = i;
        flag = 1;
        aug = min(aug, e[i].va);
        u = v;
        if (u == T) {
          ans += aug;
          for (; u != S; u = e[pre[u] ^ 1].other) {
            e[pre[u]].va -= aug;
            e[pre[u] ^ 1].va += aug;
          }
          aug = inf;
        }
        break;
      }
    }
    if (flag) continue;
    if (--vh[dis[u]] == 0) break;
    int Min = T, j;
    for (int i = head[u]; i; i = e[i].next) {
      int v = e[i].other;
      if (e[i].va > 0 && dis[v] < Min) {
        Min = dis[v];
        j = i;
      }
    }
    vh[dis[u] = Min + 1]++;
    di[u] = j;
    if (u != S) {
      u = e[pre[u] ^ 1].other;
      aug = his[u];
    }
  }
}
int main() {
  read(n);
  read(m);
  S = 1;
  tot = 1;
  for (int i = 1; i <= n; i++) {
    read(p[i].a);
    read(p[i].b);
    read(p[i].c);
  }
  int cnt = 1;
  Max = 1000000;
  for (int i = 1; i <= n; i++) {
    read(p[i].l);
    read(p[i].r);
    for (int j = p[i].l - 1; j <= p[i].r; j++) num[i][j + 101] = ++cnt;
    for (int j = p[i].l; j <= p[i].r; j++)
      add(num[i][j + 101 - 1], num[i][j + 101], Max - p[i].y(j));
    add(S, num[i][p[i].l - 1 + 101], inf);
  }
  T = ++cnt;
  for (int i = 1; i <= n; i++) add(num[i][p[i].r + 101], T, inf);
  for (int i = 1; i <= m; i++) {
    read(u);
    read(v);
    read(d);
    for (int j = p[u].l - 1; j <= p[u].r; j++)
      if (j - d >= p[v].l - 1 && j - d <= p[v].r)
        add(num[u][j + 101], num[v][j - d + 101], inf);
  }
  flow();
  ans = Max * n - ans;
  printf("%d\n", ans);
  return 0;
}