Problem: There are n cities along the road, which can be represented as a straight line. The i-th city is situated at the distance of a_i kilometers from the origin. All cities are situated in the same direction from the origin. There are m trucks travelling from one city to another. 

Each truck can be described by 4 integers: starting city s_i, finishing city f_i, fuel consumption c_i and number of possible refuelings r_i. The i-th truck will spend c_i litres of fuel per one kilometer. 

When a truck arrives in some city, it can be refueled (but refueling is impossible in the middle of nowhere). The i-th truck can be refueled at most r_i times. Each refueling makes truck's gas-tank full. All trucks start with full gas-tank.

All trucks will have gas-tanks of the same size V litres. You should find minimum possible V such that all trucks can reach their destinations without refueling more times than allowed.

Input

First line contains two integers n and m (2 ≤ n ≤ 400, 1 ≤ m ≤ 250000) — the number of cities and trucks.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9, a_i < a_{i+1}) — positions of cities in the ascending order.

Next m lines contains 4 integers each. The i-th line contains integers s_i, f_i, c_i, r_i (1 ≤ s_i < f_i ≤ n, 1 ≤ c_i ≤ 10^9, 0 ≤ r_i ≤ n) — the description of the i-th truck.

Output

Print the only integer — minimum possible size of gas-tanks V such that all trucks can reach their destinations.

Example

Input


7 6
2 5 7 10 14 15 17
1 3 10 0
1 7 12 7
4 5 13 3
4 7 10 1
4 7 10 1
1 5 11 2


Output


55

Note

Let's look at queries in details: 

  1. the 1-st truck must arrive at position 7 from 2 without refuelling, so it needs gas-tank of volume at least 50. 
  2. the 2-nd truck must arrive at position 17 from 2 and can be refueled at any city (if it is on the path between starting point and ending point), so it needs gas-tank of volume at least 48. 
  3. the 3-rd truck must arrive at position 14 from 10, there is no city between, so it needs gas-tank of volume at least 52. 
  4. the 4-th truck must arrive at position 17 from 10 and can be refueled only one time: it's optimal to refuel at 5-th city (position 14) so it needs gas-tank of volume at least 40. 
  5. the 5-th truck has the same description, so it also needs gas-tank of volume at least 40. 
  6. the 6-th truck must arrive at position 14 from 2 and can be refueled two times: first time in city 2 or 3 and second time in city 4 so it needs gas-tank of volume at least 55. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
char getc() {
  char c = getchar();
  while ((c < 'A' || c > 'Z') && (c < 'a' || c > 'z') && (c < '0' || c > '9'))
    c = getchar();
  return c;
}
int gcd(int n, int m) { return m == 0 ? n : gcd(m, n % m); }
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return x * f;
}
int n, m, a[410];
long long ans;
struct data {
  int s, f, c, r;
} b[250010];
bool check(long long k, int i) {
  long long cur = k;
  int cnt = 0;
  for (int j = b[i].s; j < b[i].f; j++) {
    if (cur >= 1ll * b[i].c * (a[j + 1] - a[j]))
      cur -= 1ll * b[i].c * (a[j + 1] - a[j]);
    else {
      cur = k, cnt++;
      if (cur >= 1ll * b[i].c * (a[j + 1] - a[j]))
        cur -= 1ll * b[i].c * (a[j + 1] - a[j]);
      else
        return 0;
    }
    if (cnt > b[i].r) return 0;
  }
  return 1;
}
signed main() {
  srand(time(0));
  n = read(), m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= m; i++)
    b[i].s = read(), b[i].f = read(), b[i].c = read(), b[i].r = read();
  random_shuffle(b + 1, b + m + 1);
  for (int i = 1; i <= m; i++)
    if (!check(ans, i)) {
      long long l = ans + 1, r = 1000000000000000000ll;
      while (l <= r) {
        long long mid = l + r >> 1;
        if (check(mid, i))
          ans = mid, r = mid - 1;
        else
          l = mid + 1;
      }
    }
  cout << ans;
  return 0;
}