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;
inline void read(int &x) {
  x = 0;
  char ch;
  bool flag = false;
  while (ch = getchar(), ch < '!')
    ;
  if (ch == '-') ch = getchar(), flag = true;
  while (x = 10 * x + ch - '0', ch = getchar(), ch > '!')
    ;
  if (flag) x = -x;
}
inline int cat_max(const int &a, const int &b) { return a > b ? a : b; }
inline int cat_min(const int &a, const int &b) { return a < b ? a : b; }
const int maxn = 51;
const int maxm = 128;
const int maxnode = 25400;
const int maxedge = 254000;
const int inf = 0x3f3f3f3f;
struct Edge {
  int to, next, cap;
} G[maxedge];
int head[maxnode], cnt = 1;
void add(int u, int v, int c) {
  G[++cnt].to = v;
  G[cnt].next = head[u];
  G[cnt].cap = c;
  head[u] = cnt;
}
inline void insert(int u, int v, int c) {
  add(u, v, c);
  add(v, u, 0);
}
int dis[maxnode], q[maxnode], l, r;
int S, T;
bool bfs() {
  memset(dis, -1, sizeof dis);
  dis[S] = 0;
  l = 0;
  r = -1;
  q[++r] = S;
  while (l <= r) {
    int u = q[l++];
    for (int i = head[u]; i; i = G[i].next) {
      if (dis[G[i].to] == -1 && G[i].cap) {
        dis[G[i].to] = dis[u] + 1;
        q[++r] = G[i].to;
      }
    }
  }
  return dis[T] != -1;
}
int cur[maxnode];
int dfs(int u, int f) {
  if (u == T || f == 0) return f;
  int ret = 0;
  for (int &i = cur[u]; i; i = G[i].next) {
    if (dis[G[i].to] == dis[u] + 1 && G[i].cap) {
      int x = dfs(G[i].to, cat_min(G[i].cap, f));
      ret += x;
      f -= x;
      G[i].cap -= x;
      G[i ^ 1].cap += x;
      if (f == 0) break;
    }
  }
  if (ret == 0) dis[u] = -1;
  return ret;
}
inline int dinic() {
  int ret = 0;
  while (bfs()) {
    memcpy(cur, head, sizeof head);
    ret += dfs(S, inf);
  }
  return ret;
}
int n;
inline int f(int x, int y) {
  y += 102;
  return x * 210 + y;
}
int a[maxn], b[maxn], c[maxn], le[maxn], ri[maxn];
inline int calc(int i, int x) { return a[i] * x * x + b[i] * x + c[i]; }
int main() {
  S = maxnode - 5;
  T = S + 1;
  read(n);
  int m;
  read(m);
  int lim = -inf;
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    read(b[i]);
    read(c[i]);
  }
  for (int i = 1; i <= n; ++i) {
    read(le[i]);
    read(ri[i]);
    insert(S, f(i, le[i]), inf);
    for (int j = le[i]; j <= ri[i]; ++j) {
      lim = cat_max(lim, calc(i, j) + 1);
    }
    insert(f(i, ri[i] + 1), T, inf);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = le[i]; j <= ri[i]; ++j) {
      insert(f(i, j), f(i, j + 1), lim - calc(i, j));
    }
  }
  for (int i = 1, u, v, d; i <= m; ++i) {
    read(u);
    read(v);
    read(d);
    for (int j = le[u]; j <= ri[u]; ++j) {
      if (j - d >= le[v] && j - d <= ri[v] + 1)
        insert(f(u, j), f(v, j - d), inf);
    }
  }
  int ans = dinic();
  printf("%d\n", lim * n - ans);
  getchar();
  getchar();
  return 0;
}