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;
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  return out << '(' << p.first << ", " << p.second << ')';
}
vector<int> v;
vector<vector<pair<int, pair<int, int>>>> q;
int dp[401][400] = {};
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  srand(time(0));
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    v.push_back(a);
  }
  q.resize(n);
  for (int i = 0; i < m; i++) {
    int s, f, c, r;
    cin >> s >> f >> c >> r;
    q[s - 1].push_back({f - 1, {c, r}});
  }
  int64_t ans = 0;
  for (int l = 0; l < n - 1; l++) {
    for (int i = 0; i <= n; i++)
      for (int j = 0; j < n; j++) dp[i][j] = 0;
    for (int t = 0; t <= n; t++) {
      for (int r = l + 1, rr = l; r < n; r++) {
        if (!t)
          dp[t][r] = v[r] - v[l];
        else {
          while (rr + 1 < r && v[r] - v[rr + 1] > dp[t - 1][rr + 1]) rr++;
          dp[t][r] = min(max(dp[t - 1][rr], v[r] - v[rr]),
                         max(dp[t - 1][rr + 1], v[r] - v[rr + 1]));
        }
      }
    }
    for (auto x : q[l])
      ans = max(ans, dp[x.second.second][x.first] * (int64_t)x.second.first);
  }
  cout << ans << endl;
}