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 mod = 1000000007;
const long long INF = (long long)1000000007 * 1000000007;
const long double eps = 1e-8;
const long double pi = acos(-1.0);
int n, m;
int a[400];
int need[400][400][400];
int mem[401];
void init() {
  for (int i = 0; i < 400; i++) {
    for (int j = 0; j < 400; j++) {
      for (int k = 0; k < 400; k++) {
        need[i][j][k] = mod;
      }
    }
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  init();
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 401; j++) {
      mem[j] = i + j;
    }
    need[i][i][0] = 0;
    for (int j = i + 1; j < n; j++) {
      for (int k = j - i; k >= 1; k--) {
        while (mem[k - 1] + 1 < j &&
               need[i][mem[k - 1] + 1][k - 1] < a[j] - a[mem[k - 1] + 1]) {
          mem[k - 1]++;
        }
        int id = mem[k - 1];
        need[i][j][k] =
            min(need[i][j][k], max(need[i][id][k - 1], a[j] - a[id]));
        if (id + 1 < j) {
          need[i][j][k] =
              min(need[i][j][k], max(need[i][id + 1][k - 1], a[j] - a[id + 1]));
        }
      }
      need[i][j][0] = a[j] - a[i];
    }
  }
  long long ans = 0;
  for (int i = 0; i < m; i++) {
    int l, r, c, d;
    cin >> l >> r >> c >> d;
    d = min(d, r - l);
    l--;
    r--;
    ans = max(ans, (long long)c * need[l][r][d]);
  }
  cout << ans << endl;
  return 0;
}