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 maxn = 55, maxp = maxn * 200, inf = 0x3f3f3f3f;
struct Edge {
  int next, to, cap, flow;
  Edge(int a = 0, int b = 0, int c = 0) : next(a), to(b), cap(c), flow(0) {}
};
vector<Edge> e(2);
int head[maxp];
void Add(int u, int v, int c) {
  e.push_back(Edge(head[u], v, c));
  head[u] = e.size() - 1;
  e.push_back(Edge(head[v], u, 0));
  head[v] = e.size() - 1;
}
int N, M, a[maxn], b[maxn], c[maxn], l[maxn], r[maxn];
int id[maxn][205];
inline int f(int i, int j) { return a[i] * j * j + b[i] * j + c[i]; }
int src, sink, cnt, cur[maxp], d[maxp], vis[maxp], T;
queue<int> q;
bool bfs() {
  vis[src] = ++T;
  q.push(src);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i; i = e[i].next) {
      int to = e[i].to;
      if (vis[to] != T && e[i].cap > e[i].flow) {
        vis[to] = T;
        d[to] = d[x] + 1;
        q.push(to);
      }
    }
  }
  return vis[sink] == T;
}
int dfs(int x, int a) {
  if (x == sink || !a) return a;
  int flow = 0, f;
  for (int &i = cur[x]; i; i = e[i].next) {
    int to = e[i].to;
    if (d[to] == d[x] + 1 && (f = dfs(to, min(a, e[i].cap - e[i].flow))) > 0) {
      e[i].flow += f;
      e[i ^ 1].flow -= f;
      flow += f;
      a -= f;
      if (!a) break;
    }
  }
  return flow;
}
int dinic() {
  int flow = 0;
  while (bfs()) {
    memcpy(cur, head, sizeof cur);
    flow += dfs(src, inf);
  }
  return flow;
}
int main() {
  scanf("%d%d", &N, &M);
  src = 1, sink = 2, cnt = 2;
  int ans = 0;
  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 fm = -inf;
    for (int j = l[i]; j <= r[i]; ++j) fm = max(fm, f(i, j));
    ans += fm;
    for (int j = l[i]; j <= r[i]; ++j) {
      id[i][j + 100] = ++cnt;
      if (j == l[i])
        Add(src, cnt, fm - f(i, j));
      else
        Add(cnt - 1, cnt, fm - f(i, j));
    }
    Add(cnt, sink, inf);
  }
  for (int i = 1, u, v, d; i <= M; ++i) {
    scanf("%d%d%d", &u, &v, &d);
    for (int j = l[v]; j <= r[v]; ++j) {
      int k = j + d;
      if (k > r[u]) break;
      if (l[u] <= k && k <= r[u]) {
        Add(id[u][k + 100], id[v][j + 100], inf);
      } else {
        Add(src, id[v][j + 100], inf);
      }
    }
  }
  printf("%d\n", ans - dinic());
  getchar(), getchar();
  return 0;
}