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;
const long long MAX_V = 1e18 + 1;
const int MAXTRUCK = 250000;
const int MAXN = 400;
struct TRUCK {
  int s, f;
  int r;
  long long c;
  friend istream& operator>>(istream& in, TRUCK& yo) {
    in >> yo.s >> yo.f;
    yo.s--;
    yo.f--;
    in >> yo.c;
    in >> yo.r;
    return in;
  }
};
int n, m;
long long a[MAXN + 2];
TRUCK truck[MAXTRUCK + 2];
bool alcanza(int idx, long long V) {
  int recargas = truck[idx].r;
  long long costo = truck[idx].c;
  long long cap = V;
  for (int i = truck[idx].s; i < truck[idx].f; i++) {
    long long dist = (a[i + 1] - a[i]) * costo;
    if (dist <= cap) {
      cap -= dist;
    } else {
      if (recargas == 0) {
        return false;
      }
      cap = V;
      recargas--;
      i--;
    }
  }
  return true;
}
long long busca_min(int idx, long long izq = 0) {
  long long der = MAX_V;
  while (izq < der) {
    long long medio = (der - izq) / 2;
    medio += izq;
    if (alcanza(idx, medio)) {
      der = medio;
    } else {
      izq = medio + 1;
    }
  }
  return izq;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) cin >> truck[i];
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  shuffle(truck, truck + m, rng);
  long long V = 0;
  for (int i = 0; i < m; ++i) {
    if (alcanza(i, V)) continue;
    V = busca_min(i, V);
  }
  cout << V << '\n';
  return 0;
}