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;
int n, m;
struct camion {
  int d, h, c, r;
};
camion t[250000];
int a[401];
bool sirve(int v, int idCamion) {
  int litros = v;
  int cantRecargas = t[idCamion].r;
  for (int i = t[idCamion].d + 1; i <= t[idCamion].h; i++) {
    if (v < (a[i] - a[i - 1])) {
      return false;
    }
    if (litros < (a[i] - a[i - 1])) {
      if (cantRecargas) {
        litros = v;
        cantRecargas--;
      } else {
        return false;
      }
    }
    litros -= (a[i] - a[i - 1]);
  }
  return true;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &a[i + 1]);
  for (int i = 0; i < m; i++)
    scanf("%d %d %d %d", &t[i].d, &t[i].h, &t[i].c, &t[i].r);
  set<int> dif;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) dif.insert(a[j] - a[i]);
  vector<int> diferencias;
  diferencias.reserve(dif.size());
  for (auto i : dif) diferencias.push_back(i);
  long long minV = 0;
  for (int i = 0; i < m; i++) {
    int ini = 0;
    int fin = diferencias.size() - 1;
    while (ini + 1 < fin) {
      int m = (ini + fin) / 2;
      if (sirve(diferencias[m], i)) {
        fin = m;
      } else {
        ini = m;
      }
    }
    if (sirve(diferencias[ini], i)) {
      minV = max(minV, ((long long)diferencias[ini]) * ((long long)t[i].c));
    } else {
      minV = max(minV, ((long long)diferencias[fin]) * ((long long)t[i].c));
    }
  }
  printf("%I64d\n", minV);
  return 0;
}