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;
long long int a[300009];
vector<pair<pair<long long int, long long int>,
            pair<long long int, long long int>>>
    v;
long long int ans = 1;
vector<int> vec;
bool check(int idx, long long int mid) {
  long long int dist, i, j, x, fuel = mid, cnt = v[idx].second.second,
                               kitna_chlegi;
  for (i = v[idx].first.first; i < v[idx].first.second; i++) {
    x = a[i + 1] - a[i];
    kitna_chlegi = fuel / v[idx].second.first;
    if (kitna_chlegi < x) {
      if (cnt == 0 || mid / v[idx].second.first < x) {
        return 0;
      }
      fuel = mid;
      cnt--;
    }
    fuel -= x * v[idx].second.first;
  }
  return 1;
}
void update(int idx) {
  if (check(idx, ans)) {
    return;
  }
  long long int l = ans, r = 1000000000000000000, temp = r;
  bool f;
  while (l <= r) {
    long long int mid = (l + r) / 2;
    f = check(idx, mid);
    if (f) {
      temp = min(temp, mid);
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  ans = max(ans, temp);
}
int main() {
  long long int i, j, k, x, y, t, n, m, p;
  scanf("%lld%lld", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  for (i = 0; i < m; i++) {
    scanf("%lld%lld", &x, &y);
    scanf("%lld%lld", &j, &k);
    v.push_back(make_pair(make_pair(x, y), make_pair(j, k)));
    vec.push_back(i);
  }
  random_shuffle(vec.begin(), vec.end());
  for (i = 0; i < m; i++) {
    update(vec[i]);
  }
  cout << ans << endl;
  ;
}