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 maxn = 2e5 + 7;
const int mod = 1e9 + 7;
const long long inf = 1e16;
const long long kk = 1ll << 50;
int a[100], b[100], c[100];
int p[100][300];
int l[100], r[100];
struct DINIC {
  const long long inf = (long long)1e16;
  int dep[20000], head[20000], cur[20000], tot;
  struct EDGE {
    int to, nxt;
    long long val;
  } g[(int)1e6 + 10];
  void add(int u, int v, long long w) {
    g[tot] = {v, head[u], w};
    head[u] = tot++;
  }
  void init() {
    tot = 0;
    memset(head, -1, sizeof(head));
  }
  void addedge(int u, int v, long long w) {
    add(u, v, w);
    add(v, u, 0);
  }
  bool bfs(int s, int t) {
    memset(dep, 0, sizeof(dep));
    queue<int> Q;
    Q.push(s);
    dep[s] = 1;
    cur[s] = head[s];
    while (!Q.empty()) {
      s = Q.front();
      Q.pop();
      for (int i = head[s], v; ~i; i = g[i].nxt) {
        if (g[i].val > 0 && !dep[v = g[i].to]) {
          dep[v] = dep[s] + 1;
          cur[v] = head[v];
          if (v == t) return true;
          Q.push(v);
        }
      }
    }
    return false;
  }
  long long dfs(int s, int t, long long flow) {
    if (s == t || flow <= 0) return flow;
    long long rest = flow;
    for (int &i = cur[s]; ~i; i = g[i].nxt) {
      int v = g[i].to;
      if (g[i].val > 0 && dep[v] == dep[s] + 1) {
        long long tmp = dfs(v, t, min(rest, g[i].val));
        if (tmp <= 0) dep[v] = 0;
        rest -= tmp;
        g[i].val -= tmp;
        g[i ^ 1].val += tmp;
        if (rest <= 0) break;
      }
    }
    return flow - rest;
  }
  long long maxflow(int s, int t) {
    long long ans = 0;
    while (bfs(s, t)) ans += dfs(s, t, inf);
    return ans;
  }
} dinic;
void rua() {
  int n, m, cnt = 1;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d%d%d", &a[i], &b[i], &c[i]);
  dinic.init();
  int S = 0, T = 1;
  for (int i = 1; i <= n; i++) {
    dinic.addedge(S, cnt + 1, inf);
    scanf("%d%d", &l[i], &r[i]);
    for (int j = l[i]; j <= r[i]; j++) {
      p[i][j - l[i]] = ++cnt;
      int tmp = a[i] * j * j + b[i] * j + c[i];
      dinic.addedge(cnt, cnt + 1, kk - tmp);
    }
    p[i][r[i] - l[i] + 1] = ++cnt;
    dinic.addedge(cnt, T, inf);
  }
  for (int i = 1; i <= m; i++) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    for (int j = max(l[v], l[u] - d); j <= min(r[v], r[u] - d) + 1; j++)
      dinic.addedge(p[u][j - l[u] + d], p[v][j - l[v]], inf);
  }
  printf("%lld\n", kk * n - dinic.maxflow(S, T));
}
int main() {
  rua();
  return 0;
}