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>
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
const int ms = 300300;
int a[505];
int s[ms], f[ms], r[ms], p[ms];
long long c[ms];
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int n, m;
  std::cin >> n >> m;
  for (int i = 0; i < n; i++) {
    std::cin >> a[i];
  }
  for (int i = 0; i < m; i++) {
    std::cin >> s[i] >> f[i] >> c[i] >> r[i];
    p[i] = i;
  }
  std::shuffle(p, p + m, rng);
  auto simul = [&](int j, long long tot) {
    long long cur = tot;
    int got = 0;
    for (int i = s[j] - 1; i < f[j] - 1; i++) {
      if ((long long)c[j] * (a[i + 1] - a[i]) > cur) {
        cur = tot;
        got++;
      }
      if ((long long)c[j] * (a[i + 1] - a[i]) > tot) {
        return false;
      }
      cur -= c[j] * (a[i + 1] - a[i]);
    }
    return got <= r[j];
  };
  long long ans = 0;
  for (int i = 0; i < m; i++) {
    if (simul(p[i], ans)) continue;
    long long rr = 1e18;
    while (ans != rr) {
      long long mid = (ans + rr) / 2;
      if (simul(p[i], mid)) {
        rr = mid;
      } else {
        ans = mid + 1;
      }
    }
    assert(simul(p[i], ans));
  }
  std::cout << ans << '\n';
}