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 N = 401;
const int M = 250001;
int a[N], s[M], f[M], c[M], r[M];
vector<int> _s;
int remain[N][N];
int need[N][N];
int n, m;
bool cmp(int i, int j) { return s[i] < s[j]; }
int calculate_remain(int i, int j, int re);
int calculate_need(int i, int j, int re) {
  if (need[j][re] != -1) return need[j][re];
  if (i == j) return need[j][re] = 0;
  int nd;
  if (re == 0) {
    nd = a[j] - a[i];
    remain[j][re] = 0;
  } else if (calculate_remain(i, j - 1, re) >= a[j] - a[j - 1]) {
    nd = calculate_need(i, j - 1, re);
    remain[j][re] = remain[j - 1][re] - a[j] + a[j - 1];
  } else {
    int w = j - 1;
    while (w > i &&
           max(calculate_need(i, w, re - 1), a[j] - a[w]) >=
               max(calculate_need(i, w - 1, re - 1), a[j] - a[w - 1])) {
      w--;
    }
    nd = max(calculate_need(i, w, re - 1), a[j] - a[w]);
    remain[j][re] = nd - a[j] + a[w];
  }
  return need[j][re] = nd;
}
int calculate_remain(int i, int j, int re) {
  if (remain[j][re] != -1) return remain[j][re];
  if (i == j) return remain[j][re] = 0;
  calculate_need(i, j, re);
  return remain[j][re];
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  _s.push_back(0);
  for (int i = 1; i <= m; i++) {
    _s.push_back(i);
    cin >> s[i] >> f[i] >> c[i] >> r[i];
  }
  sort(_s.begin() + 1, _s.end(), cmp);
  memset(remain, -1, sizeof(remain));
  memset(need, -1, sizeof(need));
  long long ans = 0;
  for (int i = 1; i <= m; i++) {
    if (i >= 2 && s[_s[i]] != s[_s[i - 1]]) {
      memset(remain, -1, sizeof(remain));
      memset(need, -1, sizeof(need));
    }
    ans = max(ans, (long long)calculate_need(s[_s[i]], f[_s[i]], r[_s[i]]) *
                       c[_s[i]]);
  }
  cout << ans;
  return 0;
}