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 INF = 1e9, MAX = 1e6;
struct Edge {
  int v, next, cap;
} edge[5000009];
int head[20009], h[20009], gap[20009], tot, n, src, des, N;
inline void addedge(int u, int v, int cap) {
  edge[tot].v = v;
  edge[tot].cap = cap;
  edge[tot].next = head[u];
  head[u] = tot++;
  edge[tot].v = u;
  edge[tot].cap = 0;
  edge[tot].next = head[v];
  head[v] = tot++;
}
inline int dfs(int u, int cap) {
  if (u == des) return cap;
  int minh = n - 1;
  int lv = cap, d;
  for (int e = head[u]; e != -1; e = edge[e].next) {
    int v = edge[e].v;
    int w = edge[e].cap;
    if (w > 0) {
      if (h[v] + 1 == h[u]) {
        d = min(lv, edge[e].cap);
        d = dfs(v, d);
        edge[e].cap -= d;
        edge[e ^ 1].cap += d;
        lv -= d;
        if (h[src] >= n) return cap - lv;
        if (lv == 0) break;
      }
      minh = min(minh, h[v]);
    }
  }
  if (lv == cap) {
    --gap[h[u]];
    if (gap[h[u]] == 0) h[src] = n;
    h[u] = minh + 1;
    ++gap[h[u]];
  }
  return cap - lv;
}
int sap() {
  int res = 0;
  memset(gap, 0, sizeof(gap));
  memset(h, 0, sizeof(h));
  gap[0] = n;
  while (h[src] < n) res += dfs(src, INF);
  return res;
}
int a[20009], b[20009], c[20009], L[20009], R[20009];
const int MV = 120;
int id(int x, int y) { return x * 300 + y; }
int main() {
  memset(head, -1, sizeof(head));
  tot = 0;
  int N, M;
  scanf("%d%d", &N, &M);
  src = N * 300 + 500;
  des = src + 1;
  n = des + 1;
  for (int i = 0; i < N; i++) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
  }
  for (int i = 0; i < N; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    L[i] = l;
    R[i] = r;
    addedge(src, id(i, l - 1 + MV), INF);
    for (int j = l; j <= r; j++)
      addedge(id(i, j + MV - 1), id(i, j + MV),
              MAX - a[i] * j * j - b[i] * j - c[i]);
    addedge(id(i, r + MV), des, INF);
  }
  for (int i = 0; i < M; i++) {
    int u, v, d;
    scanf("%d%d%d", &u, &v, &d);
    u--;
    v--;
    for (int j = L[u] - 1; j <= R[u]; j++) {
      int val = j - d;
      if (L[v] - 1 <= val && val <= R[v]) {
        addedge(id(u, j + MV), id(v, val + MV), INF);
      }
    }
  }
  long long ans = sap();
  ans = 1LL * N * MAX - ans;
  printf("%I64d\n", ans);
  return 0;
}