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 node {
  int s, f, c, r;
};
vector<node> vec[420];
pair<int, int> st[420];
int f[420][420];
int n, m, a[420], tops, mx[420][420];
long long ans;
int main() {
  scanf("%d %d", &n, &m);
  for (register int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (register int i = 1; i <= n; i++)
    for (register int j = i + 1; j <= n; j++) {
      mx[i][j] = max(mx[i][j - 1], a[j] - a[j - 1]);
    }
  for (register int i = 1; i <= m; i++) {
    int s, f, c, r;
    scanf("%d %d %d %d", &s, &f, &c, &r);
    vec[s].push_back((node){s, f, c, r});
  }
  for (register int s = 1; s <= n; s++) {
    tops = 0;
    for (register int i = s; i <= n; i++) f[i][0] = a[i] - a[s];
    for (register int k = 1; k <= n - s; k++) {
      int id = s;
      for (register int i = s + 1; i <= n; i++) {
        while (id < i - 1 && max(f[id][k - 1], a[i] - a[id]) >
                                 max(f[id + 1][k - 1], a[i] - a[id + 1]))
          ++id;
        f[i][k] = max(f[id][k - 1], a[i] - a[id]);
      }
    }
    for (register int i = 0; i < (int)vec[s].size(); i++) {
      node cur = vec[s][i];
      ans = max(ans, (long long)cur.c * f[cur.f][min(cur.r, n - s)]);
    }
  }
  cout << ans << endl;
}