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 Tedge {
  int to, next;
  long long f;
} edges[500000];
const int maxn = 100000;
int head[maxn], bs = -1;
void add(int now, int to, long long f) {
  bs++;
  edges[bs].to = to;
  edges[bs].f = f;
  edges[bs].next = head[now];
  head[now] = bs;
}
void addedge(int now, int to, long long f) {
  add(now, to, f);
  add(to, now, 0);
}
int n, m;
long long a[55], b[55], c[55];
int getp(int i, int x) {
  x += 100;
  return (i - 1) * 203 + x;
}
const long long big = 10000000000ll, oo = 999999999999999ll;
long long f(int i, long long x) {
  return big - (a[i] * x * x + b[i] * x + c[i]);
}
int S, T;
int L[55], R[55];
int dis[maxn];
queue<int> que;
bool bfs() {
  memset(dis, 0, sizeof(dis));
  dis[S] = 1;
  que.push(S);
  while (!que.empty()) {
    int now = que.front();
    for (int i = head[now]; i != -1; i = edges[i].next) {
      int to = edges[i].to;
      if (!edges[i].f || dis[to]) continue;
      dis[to] = dis[now] + 1;
      que.push(to);
    }
    que.pop();
  }
  return dis[T];
}
long long dfs(int now, long long maxf) {
  if (now == T) return maxf;
  long long t = 0, f;
  for (int i = head[now]; i != -1; i = edges[i].next) {
    int to = edges[i].to;
    if (!edges[i].f || (dis[now] + 1 != dis[to])) continue;
    f = dfs(to, min(maxf, edges[i].f));
    t += f;
    maxf -= f;
    edges[i].f -= f;
    edges[i ^ 1].f += f;
    if (!maxf) break;
  }
  if (!t) dis[now] = -9999;
  return t;
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lld%lld%lld", a + i, b + i, c + i);
  S = getp(n + 1, -100);
  T = S + 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", L + i, R + i);
    addedge(S, getp(i, L[i]), oo);
    for (int j = L[i]; j <= R[i]; j++)
      addedge(getp(i, j), getp(i, j + 1), f(i, j));
    addedge(getp(i, R[i] + 1), T, oo);
  }
  for (; m; m--) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    for (int j = L[a]; j <= R[a]; j++) {
      int t = j - c;
      if (t < L[b]) t = L[b];
      if (t > R[b]) {
        addedge(getp(a, j), T, oo);
        break;
      }
      addedge(getp(a, j), getp(b, t), oo);
    }
  }
  long long ans = big * n;
  while (bfs()) ans -= dfs(S, oo);
  printf("%lld", ans);
  return 0;
}