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>
#pragma comment(linker, "/stack:16777216")
using namespace std;
const int INF = 1000000000;
const int MAX = 100000;
const int MAX2 = 100000;
const int BASE = 1000000000;
const int LIM = 2000000;
struct edge {
  int a, b;
  int c, f;
  int rev;
  edge() {}
  edge(int _a, int _b, int _c, int _f, int _rev) {
    a = _a;
    b = _b;
    c = _c;
    f = _f;
    rev = _rev;
  }
};
int n, m, N, cnt;
int fA[MAX];
int fB[MAX];
int fC[MAX];
int L[MAX];
int R[MAX];
int T[MAX];
int W[MAX];
int Start[MAX2];
int Next[MAX2];
int P[MAX2];
int C[MAX2];
edge E[MAX2];
int getVal(int i, int x) { return fA[i] * x * x + fB[i] * x + fC[i]; }
void add_edge(int u, int v, int c) {
  E[cnt] = edge(u, v, c, 0, cnt + 1);
  Next[cnt] = Start[u];
  Start[u] = cnt;
  ++cnt;
  E[cnt] = edge(v, u, 0, 0, cnt - 1);
  Next[cnt] = Start[v];
  Start[v] = cnt;
  ++cnt;
}
int Flow(int s, int t) {
  int res = 0;
  while (1) {
    queue<int> Q;
    for (int i = (0); i < (N); ++i) P[i] = -1;
    Q.push(s);
    while (!Q.empty()) {
      int cur = Q.front();
      Q.pop();
      for (int i = Start[cur]; i != -1; i = Next[i]) {
        int to = E[i].b;
        if (P[to] == -1 && to != s && E[i].c - E[i].f > 0) {
          P[to] = i;
          Q.push(to);
        }
      }
    }
    if (P[t] == -1) break;
    int val = INF, x;
    x = t;
    while (x != s) {
      int id = P[x];
      val = min(val, E[id].c - E[id].f);
      x = E[id].a;
    }
    res += val;
    x = t;
    while (x != s) {
      int id = P[x];
      E[id].f += val;
      E[E[id].rev].f -= val;
      x = E[id].a;
    }
  }
  return res;
}
int main() {
  memset(Start, -1, sizeof(Start));
  scanf("%d %d", &n, &m);
  for (int i = (0); i < (n); ++i) {
    scanf("%d %d %d", &fA[i], &fB[i], &fC[i]);
  }
  for (int i = (0); i < (n); ++i) scanf("%d %d", &L[i], &R[i]);
  int sum = 0;
  for (int i = (0); i < (n); ++i) {
    T[i] = sum;
    W[i] = R[i] - L[i] + 2;
    sum += W[i];
  }
  for (int i = (0); i < (n); ++i)
    for (int j = (1); j < (W[i]); ++j)
      add_edge(T[i] + j - 1, T[i] + j, LIM - getVal(i, L[i] + j - 1));
  for (int i = (0); i < (m); ++i) {
    int u, v, d;
    scanf("%d %d %d", &u, &v, &d);
    --u;
    --v;
    for (int j = (0); j < (W[u] - 1); ++j) {
      int a = T[u] + j;
      if (L[u] + j - d >= L[v] && L[u] + j - d <= R[v] + 1) {
        int b = T[v] + ((L[u] + j - d) - L[v]);
        add_edge(a, b, INF);
      }
    }
  }
  N = sum;
  int s = N;
  int t = N + 1;
  N += 2;
  for (int i = (0); i < (n); ++i) add_edge(s, T[i], INF);
  for (int i = (0); i < (n); ++i) add_edge(T[i] + W[i] - 1, t, INF);
  int f = Flow(s, t);
  int res = LIM * n - f;
  printf("%d\n", res);
  return 0;
}