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;
typedef priority_queue<long long, vector<long long>, greater<long long>>
    minHeap;
const int INF = 1e9 + 9;
const long long LINF = 1e17 + 9;
const long long MD = 998244353;
inline long long po(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % MD;
    a = (a * a) % MD;
    b /= 2;
  }
  return ans;
}
const int N = 3e5 + 33;
long long n, m, k, t;
int arr[402];
string s;
int dp[402][402][402];
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  cout << fixed;
  cout << setprecision(7);
  cerr << "N"
       << " is " << N << endl;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int l = 0; l < n; l++) {
    for (int r = l + 1; r < n; r++) {
      dp[l][r][0] = arr[r] - arr[l];
    }
  }
  for (int i = 1; i < n + 1; i++) {
    for (int l = 0; l < n; l++) {
      int j = l;
      for (int r = l + 1; r < n; r++) {
        int ans = max((int)dp[l][j][i - 1], arr[r] - arr[j]);
        while (j + 1 < n &&
               max((int)dp[l][j + 1][i - 1], arr[r] - arr[j + 1]) <= ans) {
          j++;
          ans = max((int)dp[l][j][i - 1], arr[r] - arr[j]);
        }
        dp[l][r][i] = ans;
      }
    }
  }
  long long ans = 0;
  while (m--) {
    int l, r, c, d;
    cin >> l >> r >> c >> d;
    ans = max(ans, 1ll * c * dp[l - 1][r - 1][d]);
  }
  cout << ans;
  return 0;
}