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 = 11000;
const int Inf = 1e8;
const int maxn = 1e7;
struct edge {
  int u, c, nxt;
} e[5000000];
int cnt, head[N], vis[N], s = N - 1, t = N - 2;
void addedge(int u, int v, int c) {
  e[cnt] = (edge){v, c, head[u]}, head[u] = cnt++;
  e[cnt] = (edge){u, 0, head[v]}, head[v] = cnt++;
}
bool bfs() {
  memset(vis, -1, sizeof vis);
  queue<int> q;
  q.push(s);
  vis[s] = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = head[u]; ~i; i = e[i].nxt) {
      int w = e[i].u;
      if (vis[w] == -1 && e[i].c > 0) {
        vis[w] = vis[u] + 1;
        q.push(w);
      }
    }
  }
  return vis[t] != -1;
}
long long findf(int x, int flow) {
  if (x == t) return (long long)flow;
  long long res = 0;
  for (int i = head[x]; ~i; i = e[i].nxt) {
    int w = e[i].u;
    if (vis[w] == vis[x] + 1 && e[i].c > 0) {
      int ff = findf(w, min(e[i].c, flow));
      if (!ff) continue;
      e[i].c -= ff;
      e[i ^ 1].c += ff;
      res += ff;
      flow -= ff;
    }
  }
  if (!res) vis[x] = -1;
  return res;
}
long long ans;
void dinic() {
  while (bfs()) ans += findf(s, Inf);
}
map<pair<int, int>, int> id;
int a[N], b[N], c[N], _, l[N], r[N];
int getid(pair<int, int> x) {
  if (id.count(x)) return id[x];
  return id[x] = ++_;
}
int cal(const int &i, const int &x) { return a[i] * x * x + b[i] * x + c[i]; }
int main() {
  memset(head, -1, sizeof head);
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = (1); i < (n + 1); ++i) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
  }
  for (int i = (1); i < (n + 1); ++i) {
    scanf("%d%d", &l[i], &r[i]);
    addedge(s, getid(make_pair(i, l[i])), Inf);
    for (int j = (l[i]); j < (r[i]); ++j)
      addedge(getid(make_pair(i, j)), getid(make_pair(i, j + 1)),
              maxn - cal(i, j));
    addedge(getid(make_pair(i, r[i])), t, maxn - cal(i, r[i]));
  }
  for (int i = (0); i < (m); ++i) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    for (int j = (l[u]); j < (r[u] + 1); ++j) {
      int k = j - d;
      if (k > r[v]) {
        addedge(getid(make_pair(u, j)), t, Inf);
        continue;
      }
      if (k < l[v]) k = l[v];
      addedge(getid(make_pair(u, j)), getid(make_pair(v, k)), Inf);
    }
  }
  dinic();
  ans = 1ll * maxn * n - ans;
  printf("%lld\n", ans);
}