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>
inline int in() {
  int k = 0, p = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') p = 0, ch = getchar();
  while (isdigit(ch)) k = k * 10 + ch - '0', ch = getchar();
  return p ? k : -k;
}
inline int min(const int &a, const int &b) { return a < b ? a : b; }
inline int max(const int &a, const int &b) { return a > b ? a : b; }
const int inf = 5e6, INF = 0x7fffffff;
class Graph {
 private:
  static const int N = 400005, M = 5000005;
  int s, t;
  int dis[N];
  int pre[N], pr[N], o[N], cur[N];
  int head[N], nxt[M], to[M], cap[M];
  double w[M];

 public:
  int n, cnt;
  inline void init(int nn, int ss, int tt) {
    n = nn, s = ss, t = tt;
    cnt = -1;
    memset(head, -1, sizeof(head));
  }
  inline void add_Edge(int u, int v, int cc, int ww) {
    to[++cnt] = v, cap[cnt] = cc, w[cnt] = ww, nxt[cnt] = head[u],
    head[u] = cnt;
  }
  inline void add(int u, int v, int cc, int ww = 0) {
    add_Edge(u, v, cc, ww), add_Edge(v, u, 0, -ww);
  }
  int dfs(int u, int dist) {
    if (u == t) return dist;
    for (int &i = cur[u]; ~i; i = nxt[i]) {
      if (dis[to[i]] == dis[u] + 1 && cap[i] != 0) {
        int d = dfs(to[i], min(dist, cap[i]));
        if (d > 0) {
          cap[i] -= d;
          cap[i ^ 1] += d;
          return d;
        }
      }
    }
    return 0;
  }
  int bfs() {
    std::queue<int> Q;
    memset(dis, 0, sizeof(dis));
    Q.push(s);
    dis[s] = 1;
    while (!Q.empty()) {
      int u = Q.front();
      Q.pop();
      for (int i = head[u]; ~i; i = nxt[i]) {
        if (cap[i] > 0 && dis[to[i]] == 0) {
          dis[to[i]] = dis[u] + 1;
          Q.push(to[i]);
        }
      }
    }
    return dis[t] > 0;
  }
  long long dinic() {
    long long res = 0, d;
    while (bfs()) {
      memcpy(cur, head, sizeof(cur));
      while ((d = dfs(s, INF))) res += d;
    }
    return res;
  }
} cx;
const int N = 55, M = 101;
int a[N], b[N], c[N], L[N], R[N];
inline int calc(int x, int i) { return a[i] * x * x + b[i] * x + c[i]; }
int main() {
  int n = in(), m = in(), s = 0, t = 201 * n + 1;
  cx.init(t + 1, s, t);
  for (int i = 1; i <= n; ++i) a[i] = in(), b[i] = in(), c[i] = in();
  for (int i = 1; i <= n; ++i) L[i] = in() + M, R[i] = in() + M;
  for (int i = 1; i <= n; ++i) {
    cx.add(s, i + (L[i] - 1) * n, INF);
    for (int j = L[i]; j <= R[i]; ++j)
      cx.add(i + (j - 1) * n, i + j * n, inf - calc(j - M, i));
    cx.add(i + R[i] * n, t, INF);
  }
  for (int i = 1; i <= m; ++i) {
    int u = in(), v = in(), d = in();
    for (int j = std::max(d, L[u] - 1); j <= R[u]; ++j)
      cx.add(u + j * n, v + (j - d) * n, INF);
  }
  long long ans = cx.dinic();
  ans -= 1ll * inf * n;
  ans = -ans;
  printf("%lld\n", ans);
  return 0;
}