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 MAX_SIZE = 50 * 210;
const int EDGE_CNT = MAX_SIZE * 2 + 100 * 300 * 3;
const int S = MAX_SIZE + 1, T = MAX_SIZE + 2;
const int BOUND = 300000;
const int INF = 0x3f3f3f3f;
const int SHIFT = 100;
int nodes[MAX_SIZE + 10], nxt[EDGE_CNT], to[EDGE_CNT], cap[EDGE_CNT];
int a[55], b[55], c[55], l[55], r[55];
int vn, en;
inline void addEdge(int f, int t, int c) {
  to[en] = t;
  cap[en] = c;
  nxt[en] = nodes[f];
  nodes[f] = en;
  ++en;
  to[en] = f;
  cap[en] = 0;
  nxt[en] = nodes[t];
  nodes[t] = en;
  ++en;
}
int n, m;
inline int calc(int a, int b, int c, int v) { return a * v * v + b * v + c; }
queue<int> q;
int dist[MAX_SIZE + 10], he[MAX_SIZE + 10];
inline int min(int a, int b) { return a < b ? a : b; }
int augment(int cur, int lim) {
  if (cur == T) return lim;
  int flow = 0;
  for (int& e = he[cur]; e != -1; e = nxt[e]) {
    if (cap[e] && dist[to[e]] == dist[cur] + 1) {
      int del = augment(to[e], min(lim, cap[e]));
      lim -= del;
      flow += del;
      cap[e] -= del;
      cap[e ^ 1] += del;
      if (!lim) break;
    }
  }
  return flow;
}
bool bfs() {
  memset(dist, -1, sizeof(dist));
  dist[S] = 0;
  q.push(S);
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    for (int e = nodes[cur]; e != -1; e = nxt[e]) {
      if (cap[e] && dist[to[e]] == -1) {
        dist[to[e]] = dist[cur] + 1;
        q.push(to[e]);
      }
    }
  }
  return dist[T] != -1;
}
int dinic() {
  int flow = 0;
  while (bfs()) {
    for (int i = 0; i < vn; i++) he[i] = nodes[i];
    he[S] = nodes[S];
    he[T] = nodes[T];
    flow += augment(S, INF);
  }
  return flow;
}
int main() {
  scanf("%d%d", &n, &m);
  vn = 201 * n;
  for (int i = 0; i < n; i++) scanf("%d%d%d", a + i, b + i, c + i);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", l + i, r + i);
    l[i] += SHIFT;
    r[i] += SHIFT;
  }
  memset(nodes, -1, sizeof(nodes));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < l[i]; j++) {
      if (j == 0)
        addEdge(S, i * 201 + j, INF);
      else
        addEdge(i * 201 + j - 1, i * 201 + j, INF);
    }
    for (int j = l[i]; j <= r[i]; j++) {
      if (j == 0)
        addEdge(S, i * 201 + j, BOUND - calc(a[i], b[i], c[i], j - SHIFT));
      else
        addEdge(i * 201 + j - 1, i * 201 + j,
                BOUND - calc(a[i], b[i], c[i], j - SHIFT));
    }
    for (int j = r[i] + 1; j < 201; j++) {
      addEdge(i * 201 + j - 1, i * 201 + j, INF);
    }
    addEdge(i * 201 + 200, T, INF);
  }
  while (m--) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    --u;
    --v;
    for (int i = 200; i >= 0; i--) {
      if (i - d >= 0 && i - d <= 200)
        addEdge(u * 201 + i, v * 201 + i - d, INF);
    }
    if (d < 0) addEdge(S, v * 201 - d - 1, INF);
  }
  printf("%d\n", BOUND * n - dinic());
  return 0;
}