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 N = 15000, M = 10000000, G = 210000, P = 110, C = 60, Inf = 20000000;
int S, T, tt = 1, cnt;
int g[C][N], l[C], r[C], a[C], b[C], c[C], v[N], lev[N], head[N], to[M], nxt[M],
    s[M];
inline int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
inline void lnk(int x, int y, int z) {
  to[++tt] = y, nxt[tt] = head[x], s[tt] = z, head[x] = tt;
  to[++tt] = x, nxt[tt] = head[y], head[y] = tt;
}
inline bool bfs() {
  memset(lev, 0, sizeof(lev));
  int t = 0, sum = 1;
  v[1] = S, lev[S] = 1;
  while (t < sum) {
    int x = v[++t];
    for (int i = head[x]; i; i = nxt[i]) {
      int y = to[i];
      if (s[i] && !lev[y]) {
        lev[y] = lev[x] + 1, v[++sum] = y;
        if (y == T) return 1;
      }
    }
  }
  return 0;
}
inline int dfs(int x, int Mx) {
  if (x == T) return Mx;
  int ret = 0;
  for (int i = head[x]; i; i = nxt[i]) {
    int y = to[i];
    if (s[i] && lev[y] == lev[x] + 1) {
      int f = dfs(y, min(Mx - ret, s[i]));
      s[i] -= f, s[i ^ 1] += f, ret += f;
      if (ret == Mx) break;
    }
  }
  if (!ret) lev[x] = 0;
  return ret;
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) a[i] = gi(), b[i] = gi(), c[i] = gi();
  S = 0, T = N - 1;
  for (int i = 1; i <= n; i++) {
    l[i] = gi(), r[i] = gi();
    lnk(S, ++cnt, Inf), g[i][l[i] - 1 + P] = cnt;
    for (int j = l[i]; j <= r[i]; j++) {
      g[i][j + P] = ++cnt;
      lnk(cnt - 1, cnt, G - a[i] * j * j - b[i] * j - c[i]);
    }
    lnk(cnt, T, Inf);
  }
  while (m--) {
    int x = gi(), y = gi(), d = gi();
    for (int i = l[x]; i <= r[x]; i++) {
      int j = i - d;
      if (j >= l[y] && j <= r[y]) lnk(g[x][i + P - 1], g[y][j + P - 1], Inf);
      if (j > r[y]) lnk(g[x][i + P - 1], g[x][i + P], Inf);
    }
  }
  int ans = G * n;
  while (bfs()) ans -= dfs(S, 2147483600);
  cout << ans;
  return 0;
}