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;
long long n, m;
long long city[402];
long long truck[250010][4];
long long idx[250010];
int check(long long i, long long val) {
  long long second = truck[i][0], first = truck[i][1], c = truck[i][2],
            r = truck[i][3];
  long long cur = val;
  for (long long j = second; j < first; j++) {
    if ((city[j + 1] - city[j]) * c > val) return 0;
    if ((city[j + 1] - city[j]) * c > cur) {
      if (r <= 0) return 0;
      r--;
      cur = val;
    }
    cur -= (city[j + 1] - city[j]) * c;
  }
  return 1;
}
long long search(long long i, long long lo, long long hi) {
  if (lo == hi) return lo;
  long long mid = lo + (hi - lo) / 2;
  if (check(i, mid))
    return search(i, lo, mid);
  else
    return search(i, mid + 1, hi);
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> city[i];
  for (int i = 1; i <= m; i++)
    cin >> truck[i][0] >> truck[i][1] >> truck[i][2] >> truck[i][3];
  for (int i = 1; i <= m; i++) idx[i] = i;
  random_shuffle(idx + 1, idx + 1 + m);
  long long cur = 0;
  for (int i = 1; i <= m; i++) {
    if (check(idx[i], cur)) continue;
    cur = search(idx[i], cur, LLONG_MAX);
  }
  std::cout << cur << '\n';
  return 0;
}