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;
struct truck {
  int s, f, c, r;
};
vector<int> a;
bool can(truck& t, long long& v) {
  long long curr, rf = 0, fu = v;
  for (int i = t.s; i < t.f; i++) {
    curr = 1LL * (a[i] - a[i - 1]) * t.c;
    rf += (curr > fu);
    fu = (fu >= curr ? fu : v) - curr;
    if (fu < 0 || rf > t.r) return false;
  }
  return true;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  a.resize(n);
  for (int& i : a) scanf("%d", &i);
  vector<truck> b(m);
  for (truck& t : b) scanf("%d%d%d%d", &t.s, &t.f, &t.c, &t.r);
  sort(b.begin(), b.end(), [&](truck& i, truck& j) {
    return 1.0 * (a[i.f - 1] - a[i.s - 1]) * i.c * j.r >
           1.0 * (a[j.f - 1] - a[j.s - 1]) * j.c * i.r;
  });
  long long v = 0;
  for (truck& t : b)
    if (!can(t, v)) {
      long long st = v + 1, en = 1e18, mid;
      while (st <= en)
        can(t, mid = (st + en) >> 1) ? en = mid - 1 : st = mid + 1;
      v = st;
    }
  printf("%I64d\n", v);
}