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 long long N = 250001;
struct driver {
  long long s, f, c, r;
};
long long n, m, a[N];
driver d[N];
bool check(long long pos, long long V) {
  if (d[pos].s < d[pos].f) {
    long long curV = V, cur = 0, cur_city = d[pos].s;
    while (cur_city != d[pos].f) {
      if (curV - (a[cur_city + 1] - a[cur_city]) * d[pos].c >= 0)
        curV -= (a[cur_city + 1] - a[cur_city]) * d[pos].c, cur_city++;
      else {
        curV = V;
        cur++;
        if (curV - (a[cur_city + 1] - a[cur_city]) * d[pos].c >= 0)
          curV -= (a[cur_city + 1] - a[cur_city]) * d[pos].c, cur_city++;
        else
          return 0;
      }
    }
    return cur <= d[pos].r;
  } else {
    long long curV = V, cur = 0, cur_city = d[pos].s;
    while (cur_city != d[pos].f) {
      if (curV - (a[cur_city] - a[cur_city - 1]) * d[pos].c >= 0)
        curV -= (a[cur_city] - a[cur_city - 1]) * d[pos].c, cur_city--;
      else {
        curV = V;
        cur++;
        if (curV - (a[cur_city] - a[cur_city - 1]) * d[pos].c >= 0)
          curV -= (a[cur_city] - a[cur_city - 1]) * d[pos].c, cur_city--;
        else
          return 0;
      }
    }
    return cur <= d[pos].r;
  }
}
signed main() {
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= m; i++) {
    long long s, f, c, r;
    cin >> s >> f >> c >> r;
    d[i] = {s, f, c, r};
  }
  random_shuffle(d + 1, d + m + 1);
  long long maxV = 0;
  for (long long i = 1; i <= m; i++) {
    if (check(i, maxV)) continue;
    long long l = maxV, r = 1000000000000000001;
    while (l < r - 1) {
      long long mm = (l + r) / 2;
      if (check(i, mm))
        r = mm;
      else
        l = mm;
    }
    maxV = r;
  }
  cout << maxV;
  return 0;
}