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 Edge {
  int ed, flow, next;
} edge[2000000];
int head[20010], nEdge;
int d[20010], pre[20010], path[20010], his[20010], cur[20010], vh[20010];
void init() {
  memset(head, -1, sizeof(head));
  nEdge = 0;
}
void addEdge(const int a, const int b, const int flow) {
  edge[nEdge].ed = b;
  edge[nEdge].flow = flow;
  edge[nEdge].next = head[a];
  head[a] = nEdge++;
  edge[nEdge].ed = a;
  edge[nEdge].flow = 0;
  edge[nEdge].next = head[b];
  head[b] = nEdge++;
}
int sap(const int v, const int src, const int dst) {
  int x, y, i, m, nowFlow = 1000000000, totalFlow = 0;
  bool flag;
  memset(d, 0, sizeof(d));
  memset(vh, 0, sizeof(vh));
  memcpy(cur, head, sizeof(cur));
  for (vh[0] = v, x = src; d[src] < v;) {
    his[x] = nowFlow;
    for (flag = false, i = cur[x]; ~i; i = edge[i].next) {
      if (edge[i].flow && d[x] == d[y = edge[i].ed] + 1) {
        flag = true;
        nowFlow = min(nowFlow, edge[i].flow);
        pre[y] = x;
        path[y] = cur[x] = i;
        if ((x = y) == dst) {
          for (totalFlow += nowFlow; x != src; x = pre[x]) {
            edge[path[x]].flow -= nowFlow;
            edge[path[x] ^ 1].flow += nowFlow;
          }
          nowFlow = 1000000000;
        }
        break;
      }
    }
    if (!flag) {
      for (m = v, i = cur[x] = head[x]; ~i; i = edge[i].next) {
        if (edge[i].flow && d[y = edge[i].ed] < m) {
          m = d[y];
          cur[x] = i;
        }
      }
      if (!(--vh[d[x]])) break;
      ++vh[d[x] = m + 1];
      if (x != src) nowFlow = his[x = pre[x]];
    }
  }
  return totalFlow;
}
int n, m, MA;
int a[100], b[100], c[100];
int l[100], r[100];
int u, v, D;
int node(int i, int x) { return i * 202 + x + 100 + 2; }
int main() {
  init();
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
  }
  MA = 200000;
  for (int i = 0; i < n; i++) {
    addEdge(0, node(i, -100), 1000000000);
    scanf("%d%d", &l[i], &r[i]);
    for (int j = -100; j <= 100; j++) {
      if ((l[i] <= j) && (j <= r[i]))
        addEdge(node(i, j), node(i, j + 1),
                MA - (a[i] * j * j + b[i] * j + c[i]));
      else
        addEdge(node(i, j), node(i, j + 1), 1000000000);
    }
    addEdge(node(i, 101), 1, 1000000000);
  }
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &u, &v, &D);
    for (int j = -100; j <= 100; j++)
      if (j - D >= -100) {
        addEdge(node(u - 1, j), node(v - 1, j - D), 1000000000);
      }
  }
  printf("%d\n", n * MA - sap(2 + n * 202, 0, 1));
}