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;
template <typename T1, typename T2>
void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
namespace fastio {
char rbuf[1 << 23], *p1 = rbuf, *p2 = rbuf, wbuf[1 << 23], *p3 = wbuf;
inline char getc() {
  return p1 == p2 &&
                 (p2 = (p1 = rbuf) + fread(rbuf, 1, 1 << 23, stdin), p1 == p2)
             ? -1
             : *p1++;
}
inline void putc(char x) { (*p3++ = x); }
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T neg = 0;
  while (!isdigit(c)) neg |= !(c ^ '-'), c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (neg) x = (~x) + 1;
}
template <typename T>
void recursive_print(T x) {
  if (!x) return;
  recursive_print(x / 10);
  putc(x % 10 ^ 48);
}
template <typename T>
void print(T x) {
  if (!x) putc('0');
  if (x < 0) putc('-'), x = ~x + 1;
  recursive_print(x);
}
void print_final() { fwrite(wbuf, 1, p3 - wbuf, stdout); }
}  // namespace fastio
const int MAXN = 50;
const int MAXV = 50 * 201 + 100;
const int MAXE = 5e4 * 2;
const int DELTA = 1e6;
const long long INF = 1e14;
int n, m, a[MAXN + 5], b[MAXN + 5], c[MAXN + 5], l[MAXN + 5], r[MAXN + 5];
int id(int x, int y) { return (x - 1) * 201 + y + 101; }
int S, T, hd[MAXV + 5], nxt[MAXE + 5], to[MAXE + 5], ec = 1;
long long cap[MAXE + 5];
void adde(int u, int v, long long f) {
  to[++ec] = v;
  cap[ec] = f;
  nxt[ec] = hd[u];
  hd[u] = ec;
  to[++ec] = u;
  cap[ec] = 0;
  nxt[ec] = hd[v];
  hd[v] = ec;
}
int dep[MAXV + 5], now[MAXV + 5];
bool getdep() {
  memset(dep, -1, sizeof(dep));
  dep[S] = 0;
  queue<int> q;
  q.push(S);
  now[S] = hd[S];
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int e = hd[x]; e; e = nxt[e]) {
      int y = to[e];
      long long z = cap[e];
      if (z && !~dep[y]) {
        dep[y] = dep[x] + 1;
        now[y] = hd[y];
        q.push(y);
      }
    }
  }
  return ~dep[T];
}
long long getflow(int x, long long f) {
  if (x == T) return f;
  long long ret = 0;
  for (int &e = now[x]; e; e = nxt[e]) {
    int y = to[e];
    long long z = cap[e];
    if (dep[y] == dep[x] + 1 && z) {
      long long w = getflow(y, min(f - ret, z));
      cap[e] -= w;
      cap[e ^ 1] += w;
      ret += w;
      if (ret == f) return ret;
    }
  }
  return ret;
}
long long dinic() {
  long long ret = 0;
  while (getdep()) ret += getflow(S, INF);
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  T = (S = 201 * n + 1) + 1;
  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]);
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    for (int j = l[u]; j <= r[u]; j++) {
      if (l[v] <= j - w && j - w <= r[v])
        adde(id(u, j), id(v, j - w), INF);
      else if (j - w > r[v])
        adde(id(u, j), T, INF);
    }
    for (int j = l[v]; j <= r[v]; j++) {
      if (j + w < l[u]) adde(S, id(v, j), INF);
    }
  }
  for (int i = 1; i <= n; i++) {
    adde(S, id(i, l[i]), DELTA - l[i] * l[i] * a[i] - l[i] * b[i] - c[i]);
    for (int j = l[i] + 1; j <= r[i]; j++)
      adde(id(i, j - 1), id(i, j), DELTA - j * j * a[i] - j * b[i] - c[i]);
    adde(id(i, r[i]), T, INF);
  }
  printf("%lld\n", n * DELTA - dinic());
  return 0;
}