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 = 52, M = 500005;
const long long INF = (long long)1ll << 50, INF2 = (long long)1ll << 30;
struct edge {
  int v, nt;
  long long w;
} e[M << 1];
void rd(int &x) {
  char c = getchar();
  x = 0;
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = x * 10 + c - 48, c = getchar();
}
int h[N * 200], cnt = 1;
inline void add(int u, int v, long long w) {
  e[++cnt] = (edge){v, h[u], w};
  h[u] = cnt;
  e[++cnt] = (edge){u, h[v], 0};
  h[v] = cnt;
}
int q1[N * 200], hd, tl;
int d[N * 200], s, t, n, m, cur[N * 200];
int bfs(int s, int t) {
  memset(d, -1, sizeof(d));
  d[s] = 0;
  hd = tl = 1;
  q1[1] = s;
  while (hd <= tl) {
    int u = q1[hd];
    for (int i = h[u]; i; i = e[i].nt) {
      int v = e[i].v;
      if (e[i].w > 0 && d[v] == -1) {
        d[v] = d[u] + 1;
        q1[++tl] = v;
      }
    }
    ++hd;
  }
  return d[t] != -1;
}
long long dinic(int u, long long a) {
  if (u == t || a == 0) return a;
  long long minf = 0, f = 0;
  for (int &i = cur[u]; i; i = e[i].nt) {
    int v = e[i].v;
    if (d[v] == d[u] + 1) {
      minf = dinic(v, min(a, e[i].w));
      if (minf > 0) {
        e[i].w -= minf;
        e[i ^ 1].w += minf;
        a -= minf;
        f += minf;
        if (!a) break;
      }
    }
  }
  return f;
}
int dd[N][210], gt = 0;
long long maxflow(int s, int t) {
  long long res = 0;
  while (bfs(s, t)) {
    for (int i = 0; i <= gt; i++) cur[i] = h[i];
    res += dinic(s, INF);
  }
  return res;
}
long long A[N], B[N], C[N];
int x, y, z;
int l[N], r[N], ct[N];
inline long long cal(int i, int x) {
  return 1ll * A[i] * x * x + 1ll * B[i] * x + C[i];
}
int main() {
  scanf("%d%d", &n, &m);
  s = 0;
  t = gt = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d%I64d%I64d", &A[i], &B[i], &C[i]);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &l[i], &r[i]);
    add(s, gt + 1, INF);
    for (int j = l[i]; j <= r[i]; j++) {
      dd[i][j - l[i]] = ++gt;
      add(gt, gt + 1, INF2 - cal(i, j));
    }
    dd[i][r[i] - l[i] + 1] = ++gt;
    add(gt, t, INF);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &x, &y, &z);
    for (int j = max(l[y], l[x] - z); j <= min(r[y], r[x] - z) + 1; j++) {
      add(dd[x][j + z - l[x]], dd[y][j - l[y]], INF);
    }
  }
  printf("%I64d", 1ll * INF2 * n - maxflow(s, t));
  return 0;
}