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 = 49, M = 209;
const int inf = 0x3f3f3f3f;
int first[N * M], cur[N * M], d[N * M], idx[N][M], a[N], b[N], c[N], l[N], r[N];
struct edg {
  int next, to, cap, flow;
} e[N * N * M * 2];
bool vis[N * M];
int n, m, lim = 100, mr = 300000, e_sum, tot, s, t;
queue<int> q;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int f(int x, int y) {
  if (y < l[x] || y > r[x]) return inf;
  return mr - (a[x] * y * y + b[x] * y + c[x]);
}
void add_edg(int x, int y, int z) {
  e_sum++;
  e[e_sum].next = first[x];
  first[x] = e_sum;
  e[e_sum].to = y;
  e[e_sum].cap = z;
  e[e_sum].flow = 0;
}
void insert(int x, int y, int z) {
  add_edg(x, y, z);
  add_edg(y, x, 0);
}
bool bfs() {
  while (!q.empty()) q.pop();
  memset(vis, 0, sizeof vis);
  q.push(s);
  vis[s] = 1;
  d[s] = 1;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = first[x]; i; i = e[i].next) {
      int w = e[i].to;
      if (e[i].cap > e[i].flow && !vis[w]) {
        q.push(w);
        vis[w] = 1;
        d[w] = d[x] + 1;
        if (w == t) return true;
      }
    }
  }
  return false;
}
int dfs(int x, int mr) {
  if (x == t || mr == 0) return mr;
  int &i = cur[x], tot = 0;
  for (; i; i = e[i].next) {
    int w = e[i].to;
    if (e[i].cap > e[i].flow && d[w] == d[x] + 1) {
      int flow = dfs(w, min(mr, e[i].cap - e[i].flow));
      if (flow) {
        tot += flow;
        mr -= flow;
        e[i].flow += flow;
        e[((i - 1) ^ 1) + 1].flow -= flow;
        if (mr == 0) return tot;
      }
    }
  }
  return tot;
}
int dinic() {
  int flow = 0;
  while (bfs()) {
    for (int i = 1; i <= tot; i++) cur[i] = first[i];
    flow += dfs(s, inf);
  }
  return flow;
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= n; i++) a[i] = read(), b[i] = read(), c[i] = read();
  for (int i = 1; i <= n; i++) l[i] = read(), r[i] = read();
  s = ++tot;
  t = ++tot;
  for (int i = 1; i <= n; i++) {
    int last = s;
    for (int j = -100; j <= 101; j++) {
      int x = j + lim;
      idx[i][x] = ++tot;
      if (j != -100)
        insert(last, idx[i][x], f(i, j - 1));
      else
        insert(last, idx[i][x], inf);
      last = idx[i][x];
    }
    insert(last, t, inf);
  }
  for (int i = 1; i <= m; i++) {
    int x = read(), y = read(), z = read();
    for (int j = -100; j <= 101; j++) {
      int p = j + lim, q = j + lim - z;
      if (q >= 0 && idx[y][q]) insert(idx[x][p], idx[y][q], inf);
    }
  }
  printf("%d\n", mr * n - dinic());
  return 0;
}