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;
vector<int> a, perm;
long long query[250005][4];
bool check(int ind, long long V) {
  long long rem = V;
  int s = query[ind][0], f = query[ind][1];
  long long r = query[ind][3], c = query[ind][2];
  while (s < f) {
    if (c * (a[s + 1] - a[s]) > V) return false;
    if (c * (a[s + 1] - a[s]) > rem) {
      rem = V;
      r--;
    }
    if (r < 0) return false;
    rem -= (c * (a[s + 1] - a[s]));
    s++;
  }
  return true;
}
long long bs(int ind) {
  long long lo = 0, hi = 4e18;
  while (lo < hi) {
    long long mi = (lo + hi) / 2ll;
    if (check(ind, mi))
      hi = mi;
    else
      lo = mi + 1ll;
  }
  return lo;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n, m;
  cin >> n >> m;
  a.resize(n);
  perm.resize(m);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) perm[i] = i;
  shuffle(perm.begin(), perm.end(), rng);
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < 4; j++) {
      cin >> query[i][j];
    }
    query[i][0]--;
    query[i][1]--;
  }
  long long ans = 0;
  for (int i = 0; i < m; i++) {
    int ind = perm[i];
    if (check(ind, ans)) continue;
    ans = bs(ind);
  }
  cout << ans << endl;
}