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 P = 1e9 + 7;
int n, m, g;
namespace flow {
int n, s, t;
int head[200006], to[200006 << 1], nex[200006 << 1], ecn = 1;
long long wto[200006 << 1];
int ade(int u, int v, int w) {
  to[++ecn] = v, nex[ecn] = head[u], wto[ecn] = w, head[u] = ecn;
  to[++ecn] = u, nex[ecn] = head[v], wto[ecn] = 0, head[v] = ecn;
  return ecn;
}
queue<int> Q;
int dep[200006], cur[200006];
const int inf = 1e9;
void clear() {
  for (int i = (1), iend = (n); i <= iend; ++i) head[i] = dep[i] = cur[i] = 0;
  ecn = 1;
}
bool bfs() {
  Q.push(s);
  for (int i = (1), iend = (n); i <= iend; ++i) cur[i] = head[i];
  memset(dep, 0x3f, sizeof dep);
  dep[s] = 0;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int i = head[u]; i; i = nex[i])
      if (dep[to[i]] > inf && wto[i]) dep[to[i]] = dep[u] + 1, Q.push(to[i]);
  }
  return dep[t] < inf;
}
long long dfs(int u, long long lim) {
  if (!lim || u == t) return lim;
  long long flow = 0, f;
  for (int& i = cur[u]; i; i = nex[i])
    if (dep[to[i]] == dep[u] + 1 && (f = dfs(to[i], min(lim, 1ll * wto[i])))) {
      lim -= f, flow += f, wto[i] -= f, wto[i ^ 1] += f;
      if (!lim) break;
    }
  return flow;
}
long long dinic() {
  long long res = 0;
  while (bfs()) res += dfs(s, 0x3f3f3f3f3f3f3f3f);
  return res;
}
}  // namespace flow
int A[200006], B[200006], C[200006], L[200006], R[200006];
int F(int i, int x) { return A[i] * x * x + B[i] * x + C[i]; }
int gt(int i, int j) { return (i - 1) * 201 + 101 + j; }
void solve() {
  cin >> n >> m;
  for (int i = (1), iend = (n); i <= iend; ++i) {
    scanf("%d%d%d", A + i, B + i, C + i);
  }
  int t = 201 * n + 3, s = 201 * n + 4;
  flow::s = s, flow::t = t, flow::n = s;
  for (int i = (1), iend = (n); i <= iend; ++i) {
    scanf("%d%d", L + i, R + i);
    for (int k = (-100), kend = (100); k <= kend; ++k) {
      int nd = gt(i, k);
      if (L[i] <= k && k < R[i]) flow::ade(nd, nd + 1, 0x3f3f3f3f - F(i, k));
      if (k == L[i]) flow::ade(s, nd, 0x7f3f3f3f);
      if (k == R[i]) flow::ade(nd, t, 0x3f3f3f3f - F(i, k));
    }
  }
  for (int i = (1), iend = (m); i <= iend; ++i) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    for (int j = (L[u]), jend = (R[u]); j <= jend; ++j) {
      int p = j - d;
      if (p >= L[v] && p <= R[v]) {
        flow::ade(gt(u, j), gt(v, p), 0x7f3f3f3f);
      } else if (p > R[v]) {
        flow::ade(gt(u, j), t, 0x7f3f3f3f);
      }
    }
  }
  cout << 0x3f3f3f3f * 1ll * n - flow::dinic() << endl;
}
signed main() { solve(); }