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 = 1 << 30;
namespace maxflow {
struct Edge {
  int v, c;
  Edge* next;
};
const int N = 2e4;
Edge* a[N];
Edge pool[10 * N];
Edge* last;
int d[N];
inline Edge* new_edge(int v, int c, Edge* next) {
  Edge* p = last++;
  p->v = v;
  p->c = c;
  p->next = next;
  return p;
}
inline void add_edge(int u, int v, int c) {
  a[u] = new_edge(v, c, a[u]);
  a[v] = new_edge(u, 0, a[v]);
}
inline Edge* dual(Edge* p) { return pool + ((p - pool) ^ 1); }
int augment(int u, int t, int bound) {
  if (u == t) return bound;
  int ret = 0;
  for (Edge* it = a[u]; it && ret < bound; it = it->next) {
    if (it->c && d[it->v] == d[u] + 1) {
      int aug = augment(it->v, t, min(it->c, bound - ret));
      ret += aug;
      it->c -= aug;
      dual(it)->c += aug;
    }
  }
  if (ret == 0) d[u] = -1;
  return ret;
}
bool path(int s, int t, int n) {
  fill(d, d + n, -1);
  d[s] = 0;
  queue<int> Q;
  Q.push(s);
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (Edge* it = a[u]; it; it = it->next) {
      if (it->c && d[it->v] == -1) {
        Q.push(it->v);
        d[it->v] = d[u] + 1;
        if (it->v == t) return true;
      }
    }
  }
  return false;
}
int dinic(int s, int t, int n) {
  int ret = 0;
  while (path(s, t, n)) {
    while (int aug = augment(s, t, INF)) {
      ret += aug;
    }
  }
  return ret;
}
}  // namespace maxflow
const int N = 50;
const int BOUND = 1e6;
const int LEN = 202;
int A[N], B[N], C[N];
int F(int x, int i) { return BOUND - (A[i] * x * x + B[i] * x + C[i]); }
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d%d", &A[i], &B[i], &C[i]);
  }
  maxflow::last = maxflow::pool;
  fill(maxflow::a, maxflow::a + LEN * n + 2, nullptr);
  int s = LEN * n, t = s + 1;
  for (int i = 0; i < n; ++i) {
    int low, high;
    scanf("%d%d", &low, &high);
    low += 101;
    high += 101;
    maxflow::add_edge(s, i * LEN, INF);
    for (int j = 1; j < LEN; ++j) {
      int cost = (j >= low && j <= high ? F(j - 101, i) : INF);
      maxflow::add_edge(i * LEN + j - 1, i * LEN + j, cost);
    }
    maxflow::add_edge(i * LEN + LEN - 1, t, INF);
  }
  while (m--) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    --x;
    --y;
    for (int i = 0; i < LEN; ++i) {
      if (i - z >= 0 && i - z < LEN) {
        maxflow::add_edge(x * LEN + i, y * LEN + i - z, INF);
      }
    }
  }
  printf("%d\n", n * BOUND - maxflow::dinic(s, t, n * LEN + 2));
  return 0;
}