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 oo = 1000000, inf = 1000000000;
const double pi = acos(-1);
int A[55], B[55], C[55], n, m;
struct Flow {
  int next[120000], e[120000], c[120000], head[12000], dis[12000], bfs[12000],
      len, s, t;
  Flow() { len = 1; }
  void reset(int source, int sink) {
    memset(head, 0, sizeof(head));
    s = source;
    t = sink;
    len = 1;
  }
  void addedge(int x, int y, int k) {
    next[++len] = head[x];
    head[x] = len;
    e[len] = y;
    c[len] = k;
    next[++len] = head[y];
    head[y] = len;
    e[len] = x;
    c[len] = 0;
  }
  bool build() {
    int l, r, i, x;
    memset(dis, 0, sizeof(dis));
    bfs[l = r = 1] = s;
    dis[s] = 1;
    while (l <= r) {
      x = bfs[l];
      for (i = head[x]; i; i = next[i])
        if (c[i] && !dis[e[i]]) {
          dis[e[i]] = dis[x] + 1;
          bfs[++r] = e[i];
        }
      l++;
    }
    return dis[t] != 0;
  }
  int dinic(int x, int v) {
    int flow = 0;
    if (x == t) return v;
    for (int i = head[x]; i; i = next[i])
      if (c[i] && dis[x] + 1 == dis[e[i]]) {
        int ff = dinic(e[i], min(v - flow, c[i]));
        c[i] -= ff;
        c[i ^ 1] += ff;
        flow += ff;
        if (flow == v) return flow;
      }
    dis[x] = -1;
    return flow;
  }
  int maxflow() {
    int ans = 0;
    while (build()) ans += dinic(s, inf);
    return ans;
  }
} flow;
int ch(int x, int y) { return (x - 1) * 210 + y + 102; }
int main() {
  scanf("%d%d", &n, &m);
  flow.reset(0, 1);
  for (int i = 1; i <= n; ++i) scanf("%d%d%d", &A[i], &B[i], &C[i]);
  for (int i = 1; i <= n; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    for (int j = -100; j <= 100; ++j) {
      int v = ((j >= l && j <= r) ? A[i] * j * j + B[i] * j + C[i] : -inf);
      flow.addedge(ch(i, j), ch(i, j + 1), -v + oo);
    }
    flow.addedge(0, ch(i, -100), inf);
    flow.addedge(ch(i, 101), 1, inf);
  }
  for (int i = 1; i <= m; ++i) {
    int x, y, k;
    scanf("%d%d%d", &x, &y, &k);
    for (int j = -100; j <= 100; ++j)
      if (j - k >= -100) flow.addedge(ch(x, j), ch(y, min(101, j - k)), inf);
  }
  printf("%d\n", oo * n - flow.maxflow());
}