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 N = 11111;
const int inf = 10000000;
struct Edge {
  int b, c, nxt;
} e[999999];
int p[N];
int th;
int pre[N], dis[N], gap[N], cur[N];
int sap(int s, int t, int n) {
  int i, v, u;
  int c, aug = -1, maxflow = 0;
  for (i = 0; i <= n; i++) cur[i] = p[i];
  memset(dis, 0, sizeof(dis));
  memset(gap, 0, sizeof(gap));
  u = pre[s] = s, gap[0] = n;
  while (dis[s] < n) {
  loop:
    for (int &ee = cur[u]; ee != -1; ee = e[ee].nxt) {
      v = e[ee].b;
      c = e[ee].c;
      if (c > 0 && dis[u] == dis[v] + 1) {
        if (aug == -1 || aug > c) aug = c;
        pre[v] = u;
        u = v;
        if (v == t) {
          maxflow += aug;
          for (u = pre[u]; v != s; v = u, u = pre[u])
            e[cur[u]].c -= aug, e[cur[u] ^ 1].c += aug;
          aug = -1;
        }
        goto loop;
      }
    }
    int mindis = n;
    for (i = p[u]; i != -1; i = e[i].nxt) {
      v = e[i].b;
      if (e[i].c > 0 && mindis > dis[v]) cur[u] = i, mindis = dis[v];
    }
    if ((--gap[dis[u]]) == 0) break;
    gap[dis[u] = mindis + 1]++;
    u = pre[u];
  }
  return maxflow;
}
void addedge(int a, int b, int c) {
  e[th].b = b;
  e[th].c = c;
  e[th].nxt = p[a];
  p[a] = th++;
  e[th].b = a;
  e[th].c = 0;
  e[th].nxt = p[b];
  p[b] = th++;
}
void init() {
  memset(p, -1, sizeof(p));
  th = 0;
}
int n, m;
int a[55], b[55], c[55], l[55], r[55], d;
int pos[55];
int find(int x, int y) { return y * y * a[x] + y * b[x] + c[x]; }
int main() {
  int S = 0, T = 0;
  init();
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i] >> b[i] >> c[i];
  for (int i = 1; i <= n; i++) cin >> l[i] >> r[i];
  for (int i = 1; i <= n; i++) {
    pos[i] = ++d;
    for (int j = l[i]; j <= r[i]; j++)
      addedge(d, d + 1, 300000 - find(i, j)), d++;
  }
  S = d + 1, T = S + 1;
  int a, b, c;
  while (m--) {
    cin >> a >> b >> c;
    for (int i = l[a]; i <= r[a]; i++) {
      int j = i - c;
      if (j <= l[b]) continue;
      if (j > r[b])
        addedge(pos[a] + i - l[a], T, inf);
      else
        addedge(pos[a] + i - l[a], pos[b] + j - l[b], inf);
    }
  }
  for (int i = 1; i <= n; i++)
    addedge(S, pos[i], inf), addedge(pos[i] + r[i] - l[i] + 1, T, inf);
  cout << n * 300000 - sap(S, T, T) << endl;
  return 0;
}