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 int Maxn = 4e2 + 7, Maxm = 25e4 + 7;
const long long Inf = 1e18;
long long n, m, ans, r[Maxm], s[Maxm], f[Maxm], c[Maxm], dp[Maxn][Maxn],
    sh[Maxm], a[Maxn];
bool valid(int x, long long y) {
  long long baghi = y, t = 0;
  for (int i = s[x]; i < f[x]; i++) {
    if ((a[i + 1] - a[i]) * c[x] > baghi)
      t++, baghi = y - (a[i + 1] - a[i]) * c[x];
    else
      baghi -= (c[x] * (a[i + 1] - a[i]));
    if ((a[i + 1] - a[i]) * c[x] > y) return false;
  }
  if (t <= r[x]) return true;
  return false;
}
long long find_ans(int x, long long ll, long long rr) {
  long long mid;
  while (rr - ll > 1) {
    mid = (ll + rr) / 2;
    if (valid(x, mid))
      rr = mid;
    else
      ll = mid;
  }
  return rr;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(0));
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++)
    cin >> s[i] >> f[i] >> c[i] >> r[i], sh[i] = i, s[i]--, f[i]--;
  random_shuffle(sh, sh + m);
  for (int i = 0; i < m; i++) {
    int now = sh[i];
    if (!valid(now, ans)) {
      ans = find_ans(now, ans, Inf);
    }
  }
  cout << ans;
}