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 MAX_N = 11111, MAX_M = 222222;
class Flow_Algorithm {
 public:
  int g[MAX_M], nxt[MAX_M], flow[MAX_M], c[MAX_N], h[MAX_N], nm;
  int d[MAX_N], q[MAX_N];
  int S, T;
  void addedge(int x, int y, int w) {
    g[nm] = y;
    nxt[nm] = c[x];
    c[x] = nm;
    flow[nm] = w;
    nm++;
  }
  void add(int x, int y, int w) {
    addedge(x, y, w);
    addedge(y, x, 0);
  }
  bool bfs() {
    memset(d, -1, sizeof d);
    d[q[1] = S] = 0;
    for (int l = 1, r = 1; l <= r; ++l) {
      int i = q[l];
      for (int k = c[i]; ~k; k = nxt[k]) {
        int j = g[k];
        if (!~d[j] && flow[k]) d[q[++r] = j] = d[i] + 1;
      }
    }
    return d[T] != -1;
  }
  int dfs(int i, int now) {
    if (!now) return 0;
    if (i == T) return now;
    int w = 0;
    for (int &k = c[i]; ~k; k = nxt[k]) {
      int j = g[k];
      if (d[j] == d[i] + 1) {
        int res;
        if (flow[k] && ((res = dfs(j, min(now - w, flow[k]))) > 0)) {
          flow[k] -= res;
          flow[k ^ 1] += res;
          w += res;
          if (w == now) return w;
        }
      }
    }
    return w;
  }
  void sets(int _S, int _T) {
    S = _S;
    T = _T;
    nm = 0;
    memset(c, -1, sizeof c);
  }
  int Dinic() {
    memcpy(h, c, sizeof h);
    int ans = 0;
    while (bfs()) {
      ans += dfs(S, 1000000000);
      memcpy(c, h, sizeof c);
    }
    return ans;
  }
} G;
int A[55], B[55], C[55], L[55], R[55];
int n, m;
int nd[55][222];
int calc(int i, int x) { return x * x * A[i] + x * B[i] + C[i]; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (1); i <= (n); ++i) scanf("%d%d%d", &A[i], &B[i], &C[i]);
  for (int i = (1); i <= (n); ++i) scanf("%d%d", &L[i], &R[i]);
  int tot = 0;
  for (int i = (1); i <= (n); ++i)
    for (int j = (L[i] - 1); j <= (R[i]); ++j) nd[i][j + 100] = ++tot;
  int S = tot + 1, T = tot + 2;
  G.sets(S, T);
  for (int i = (1); i <= (n); ++i) {
    G.add(S, nd[i][L[i] - 1 + 100], 1000000000);
    for (int j = (L[i]); j <= (R[i]); ++j)
      G.add(nd[i][j - 1 + 100], nd[i][j + 100], 20000000 - calc(i, j));
    G.add(nd[i][R[i] + 100], T, 1000000000);
  }
  for (int i = (1); i <= (m); ++i) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    for (int j = (L[u] - 1); j <= (R[u]); ++j) {
      int p = j - d, pt = 0;
      if (p < L[v])
        pt = nd[v][L[v] - 1 + 100];
      else
        pt = nd[v][p + 100];
      G.add(nd[u][j + 100], pt, 1000000000);
    }
  }
  int ans = G.Dinic();
  ans = n * 20000000 - ans;
  printf("%d\n", ans);
}