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;
static const int INF = 500000000;
template <class T>
void debug(T a, T b) {
  for (; a != b; ++a) cerr << *a << ' ';
  cerr << endl;
}
template <class T>
void chmin(T& a, const T& b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T& a, const T& b) {
  if (a < b) a = b;
}
int src, sink;
static const int MAX_SIZE = 55 * 205 + 2;
struct edge {
  int to, cap, rev;
  edge(int st, int second, int sr) {
    to = st;
    cap = second;
    rev = sr;
  }
};
vector<edge> g[MAX_SIZE];
int rankk[MAX_SIZE], seen[MAX_SIZE];
void add_edge(int from, int to, int cap) {
  g[from].push_back(edge(to, cap, g[to].size()));
  g[to].push_back(edge(from, 0, g[from].size() - 1));
}
void bfs(int v) {
  memset(rankk, -1, sizeof(rankk));
  rankk[v] = 0;
  queue<int> q;
  q.push(v);
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    for (int i = 0; i < (g[cur].size()); ++i) {
      edge& e = g[cur][i];
      if (e.cap > 0 && rankk[e.to] == -1) {
        rankk[e.to] = rankk[cur] + 1;
        q.push(e.to);
      }
    }
  }
}
int dfs(int v, int f) {
  if (v == sink) return f;
  for (int& i = seen[v]; i < g[v].size(); ++i) {
    edge& e = g[v][i];
    if (e.cap > 0 && rankk[e.to] > rankk[v]) {
      int d = dfs(e.to, min(e.cap, f));
      if (d > 0) {
        e.cap -= d;
        g[e.to][e.rev].cap += d;
        return d;
      }
    }
  }
  return 0;
}
int max_flow(int s, int t) {
  src = s;
  sink = t;
  int flow = 0;
  while (1) {
    bfs(src);
    if (rankk[sink] == -1) return flow;
    int f;
    memset(seen, 0, sizeof(seen));
    while ((f = dfs(src, INF)) > 0) {
      flow += f;
    }
  }
}
static const int LARGE = 300000;
int n, m;
int A[55], B[55], C[55], L[55], R[55];
int LEN = 202;
vector<pair<int, int> > gr[55];
int eval(int i, int j) { return A[i] * j * j + B[i] * j + C[i]; }
int main() {
  cin >> n >> m;
  for (int i = 0; i < (n); ++i) {
    cin >> A[i] >> B[i] >> C[i];
  }
  for (int i = 0; i < (n); ++i) cin >> L[i] >> R[i];
  for (int i = 0; i < (m); ++i) {
    int u, v, d;
    cin >> u >> v >> d;
    --u;
    --v;
    gr[u].push_back(make_pair(v, -d));
  }
  int S = n * LEN, T = S + 1;
  for (int i = 0; i < (n); ++i) {
    add_edge(S, i * LEN + L[i] + 100, INF);
    for (int j = L[i]; j <= R[i]; ++j) {
      add_edge(i * LEN + j + 100, i * LEN + j + 101, LARGE - eval(i, j));
    }
    for (int j = L[i] - 1; j <= R[i]; ++j) {
      for (int k = 0; k < (gr[i].size()); ++k) {
        int to = gr[i][k].first, st = max(L[to] - 1, j + gr[i][k].second);
        if (st <= R[to]) add_edge(i * LEN + j + 101, to * LEN + st + 101, INF);
      }
    }
    add_edge(i * LEN + R[i] + 101, T, INF);
  }
  int res = max_flow(S, T);
  int ans = LARGE * n - res;
  cout << ans << endl;
  return 0;
}