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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long maxn = 25e4 + 10, inf = 4e18, mod = 1e9 + 7;
const long double pi = 3.14159265359, eps = 1e-9;
struct truck {
  long long s, f, c, r;
};
long long n, m, a[maxn], id[maxn];
truck t[maxn];
long long nw, tt, fix;
bool check(long long v) {
  nw = v;
  tt = t[fix].r;
  for (int j = t[fix].s; j < t[fix].f; j++) {
    if ((a[j + 1] - a[j]) * t[fix].c > v) return 0;
    nw -= (a[j + 1] - a[j]) * t[fix].c;
    if (nw < 0) nw = v - (a[j + 1] - a[j]) * t[fix].c, tt--;
  }
  return tt >= 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(unsigned(std::time(0)));
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) {
    cin >> t[i].s >> t[i].f >> t[i].c >> t[i].r;
  }
  for (int i = 0; i < m; i++) id[i] = i;
  random_shuffle(id, id + m);
  long long l, r, mid, jj = 0;
  for (int i = 0; i < m; i++) {
    fix = id[i];
    if (check(jj)) continue;
    l = jj;
    r = inf;
    while (r - l > 1) {
      mid = (l + r) / 2;
      if (check(mid))
        r = mid;
      else
        l = mid;
    }
    jj = r;
  }
  cout << jj << endl;
  return 0;
}