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>
inline int getint() {
  register char ch;
  register bool neg = false;
  while (!isdigit(ch = getchar())) neg |= ch == '-';
  register int x = ch ^ '0';
  while (isdigit(ch = getchar())) x = (((x << 2) + x) << 1) + (ch ^ '0');
  return neg ? -x : x;
}
const int N = 51, V = 1e4 + 1, E = 1e5 + 1;
struct Edge {
  int from, to, remain, next;
};
Edge e[E];
int h[V], sz, cnt;
inline void add_edge(const int &u, const int &v, const int &w) {
  e[sz] = (Edge){u, v, w, h[u]};
  h[u] = sz++;
  e[sz] = (Edge){v, u, 0, h[v]};
  h[v] = sz++;
}
struct Func {
  int a, b, c;
  int operator()(const int &x) const { return a * x * x + b * x + c; }
};
Func f[N];
int l[N], r[N];
int lev[V], s, t, cur[V];
inline void bfs() {
  memset(lev, -1, sizeof lev);
  lev[s] = 0;
  std::queue<int> q;
  q.push(s);
  while (!q.empty()) {
    const int &x = q.front();
    for (register int i = h[x]; ~i; i = e[i].next) {
      const int &y = e[i].to, &r = e[i].remain;
      if (r && !~lev[y]) {
        lev[y] = lev[x] + 1;
        q.push(y);
      }
    }
    q.pop();
  }
}
int dfs(const int &x, const int &flow) {
  if (x == t) return flow;
  for (int &i = cur[x]; ~i; i = e[i].next) {
    const int &y = e[i].to;
    if (e[i].remain && lev[y] > lev[x]) {
      if (int d = dfs(y, std::min(flow, e[i].remain))) {
        e[i].remain -= d;
        e[i ^ 1].remain += d;
        return d;
      }
    }
  }
  return 0;
}
inline int dinic() {
  int maxflow = 0;
  for (;;) {
    bfs();
    if (!~lev[t]) break;
    memcpy(cur, h, sizeof h);
    while (int flow = dfs(s, INT_MAX)) {
      maxflow += flow;
    }
  }
  return maxflow;
}
std::map<int, int> node[N];
int main() {
  memset(h, -1, sizeof h);
  const int n = getint(), m = getint();
  for (register int i = 1; i <= n; i++) {
    const int a = getint(), b = getint(), c = getint();
    f[i] = (Func){a, b, c};
  }
  cnt = 1;
  s = cnt++;
  for (register int i = 1; i <= n; i++) {
    l[i] = getint(), r[i] = getint();
    for (register int j = l[i] - 1; j <= r[i]; j++) {
      node[i][j] = cnt++;
    }
  }
  const int M = INT_MAX / n;
  for (register int i = 1; i <= n; i++) {
    add_edge(s, node[i][l[i] - 1], INT_MAX);
    for (register int j = l[i]; j <= r[i]; j++) {
      add_edge(node[i][j - 1], node[i][j], M - f[i](j));
    }
    add_edge(node[i][r[i]], t, INT_MAX);
  }
  for (register int i = 0; i < m; i++) {
    const int u = getint(), v = getint(), d = getint();
    for (register int j = std::max(l[u], l[v] + d) - 1;
         j <= std::min(r[u], r[v] + d); j++) {
      add_edge(node[u][j], node[v][j - d], INT_MAX);
    }
  }
  printf("%d\n", M * n - dinic());
  return 0;
}