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 INF = (int)1e9;
int fox[100001], l[1001], r[1001], a[1001], b[1001], c[1001], tot, st, ed, ans,
    u, v, d, n, m, i, j, p, k;
map<pair<int, int>, int> mp;
struct Node {
  int ed, before, flow;
} s[1000001];
int que[100005], dis[100005], k1 = 1, Max = (int)1e7;
inline int bfs() {
  int i, j, p, k, l, r;
  memset(dis, -1, sizeof(dis));
  l = r = 1;
  dis[st] = 0;
  que[1] = st;
  while (l <= r) {
    p = que[l];
    for (i = fox[p]; i; i = s[i].before)
      if (s[i].flow > 0)
        if (dis[s[i].ed] == -1) {
          que[++r] = s[i].ed;
          dis[s[i].ed] = dis[p] + 1;
        }
    l++;
  }
  if (dis[ed] == -1) return 0;
  return 1;
}
void add(int p1, int p2, int p3) {
  s[++k1].ed = p2;
  s[k1].flow = p3;
  s[k1].before = fox[p1];
  fox[p1] = k1;
  s[++k1].ed = p1;
  s[k1].flow = 0;
  s[k1].before = fox[p2];
  fox[p2] = k1;
}
int dfs(int num, int flow) {
  int i, p;
  int a, nowans = 0;
  if (num == ed) return flow;
  for (i = fox[num]; i && flow; i = s[i].before)
    if (s[i].flow > 0 && dis[s[i].ed] == dis[num] + 1)
      if (a = dfs(s[i].ed, min(s[i].flow, flow))) {
        s[i].flow -= a;
        s[i ^ 1].flow += a;
        flow -= a;
        nowans += a;
      }
  if (!nowans) dis[num] = (int)1e9;
  return nowans;
}
int maxflow() {
  int i, j;
  int ans = 0;
  while (bfs()) {
    j = dfs(st, (int)1e9);
    while (j) ans += j, j = dfs(st, (int)1e9);
  }
  return ans;
}
int main() {
  st = ++tot;
  ed = ++tot;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d%d%d", &a[i], &b[i], &c[i]);
  for (i = 1; i <= n; i++) {
    int sum = -(int)1e9;
    scanf("%d%d", &l[i], &r[i]);
    for (j = l[i]; j <= r[i]; j++) {
      sum = max(sum, a[i] * j * j + b[i] * j + c[i]);
    }
    ans += sum;
    mp[make_pair(i, l[i] - 1)] = ++tot;
    add(st, tot, INF);
    for (j = l[i]; j <= r[i]; j++) {
      mp[make_pair(i, j)] = ++tot;
      add(mp[make_pair(i, j - 1)], tot, sum - a[i] * j * j - b[i] * j - c[i]);
    }
    add(tot, ed, INF);
  }
  for (i = 1; i <= m; i++) {
    scanf("%d%d%d", &u, &v, &d);
    for (j = l[u] - 1; j <= r[u]; j++)
      if (j - d >= l[v] - 1 && j - d <= r[v])
        add(mp[make_pair(u, j)], mp[make_pair(v, j - d)], INF);
  }
  printf("%d\n", ans - maxflow());
}