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;
struct Max_flow {
  int n, st, en, tot;
  int b[200005], e[200005], c[200005], pre[200005];
  int cnt[20005], cur[20005], h[20005], d[20005];
  queue<int> q;
  void init(int _n, int _st, int _en) {
    memset(h, 0, sizeof(h));
    tot = 1;
    st = _st;
    en = _en;
    n = _n;
  }
  void bfs() {
    while (!q.empty()) q.pop();
    memset(d, 10, sizeof(d));
    d[en] = 0;
    q.push(en);
    while (!q.empty()) {
      int k = q.front();
      q.pop();
      for (int go = h[k]; go; go = e[go])
        if (c[go] == 0 && d[b[go]] > d[k] + 1) {
          d[b[go]] = d[k] + 1;
          q.push(b[go]);
        }
    }
  }
  void add(int x, int y, int z) {
    b[++tot] = y;
    c[tot] = z;
    e[tot] = h[x];
    h[x] = tot;
    b[++tot] = x;
    c[tot] = 0;
    e[tot] = h[y];
    h[y] = tot;
  }
  int push_flow() {
    int x = en, flow = 168430090;
    while (x != st) {
      flow = min(flow, c[pre[x]]);
      x = b[pre[x] ^ 1];
    }
    x = en;
    while (x != st) {
      c[pre[x]] -= flow;
      c[pre[x] ^ 1] += flow;
      x = b[pre[x] ^ 1];
    }
    return flow;
  }
  int max_flow() {
    bfs();
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i < n; ++i) {
      if (d[i] != 168430090) cnt[d[i]]++;
    }
    int x = st, ans = 0;
    memset(cur, 0, sizeof(cur));
    for (int i = 0; i < n; ++i) {
      cur[i] = h[i];
    }
    while (d[st] < n) {
      if (x == en) {
        ans += push_flow();
        x = st;
      }
      int ok = 0;
      for (int go = cur[x]; go; go = e[go]) {
        if (c[go] && d[x] == d[b[go]] + 1) {
          ok = 1;
          pre[b[go]] = go;
          cur[x] = go;
          x = b[go];
          break;
        }
      }
      if (!ok) {
        int tmp = n - 1;
        for (int go = h[x]; go; go = e[go]) {
          if (c[go]) {
            tmp = min(tmp, d[b[go]]);
          }
        }
        if (--cnt[d[x]] == 0) break;
        d[x] = tmp + 1;
        cnt[d[x]]++;
        cur[x] = h[x];
        if (x != st) x = b[pre[x] ^ 1];
      }
    }
    return ans;
  }
} U;
int n, m;
int s[20005], a[20005], b[20005], c[20005], l[20005], r[20005];
int f(int i, int x) { return a[i] * x * x + b[i] * x + c[i]; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
  }
  int st = 0, en;
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &l[i], &r[i]);
    st += r[i] - l[i] + 2;
  }
  en = st + 1;
  U.init(en + 1, st, en);
  int sum = 0;
  for (int i = 0; i < n; ++i) {
    U.add(st, sum, 168430090);
    s[i] = sum;
    for (int j = 0; j < r[i] - l[i] + 1; ++j) {
      U.add(sum + j, sum + j + 1, 2000005 - f(i, l[i] + j));
    }
    sum += r[i] - l[i] + 2;
    U.add(sum - 1, en, 168430090);
  }
  int x, y, z;
  for (int i = 0; i < m; ++i) {
    scanf("%d%d%d", &x, &y, &z);
    --x;
    --y;
    for (int j = 0; j < r[x] - l[x] + 2; ++j) {
      int k = l[x] - 1 + j - z;
      if (k >= l[y] - 1 && k <= r[y]) {
        U.add(s[x] + j, s[y] + k - l[y] + 1, 168430090);
      }
    }
  }
  printf("%d\n", 2000005 * n - U.max_flow());
  return 0;
}