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;
long long int inf = 1e14;
long long int mod = 1e9 + 7;
char en = '\n';
struct obj {
  long long int s, f, c, r;
  obj(long long int s, long long int f, long long int c, long long int r)
      : s(s), f(f), c(c), r(r) {}
};
long long int dp[405][405];
long long int arr[405];
vector<obj> query[405];
long long int value(long long int currMin, long long int L, long long int R) {
  return max(currMin, arr[R] - arr[L]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, m;
  cin >> n >> m;
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
  for (long long int i = 1; i <= m; i++) {
    long long int s, f, c, r;
    cin >> s >> f >> c >> r;
    obj temp(s, f, c, min(r + 1, n));
    query[s].push_back(temp);
  }
  long long int res = 0;
  for (long long int i = 1; i <= n; i++) {
    for (long long int k = 1; k <= n; k++) {
      for (long long int j = 0; j <= n; j++) dp[k][j] = inf;
    }
    for (long long int j = i; j <= n; j++) {
      dp[1][j] = arr[j] - arr[i];
    }
    for (long long int k = 2; k <= n; k++) {
      long long int ptr = i;
      for (long long int j = i + k; j <= n; j++) {
        while (value(dp[k - 1][ptr], ptr, j) >=
               value(dp[k - 1][ptr + 1], ptr + 1, j)) {
          ptr++;
        }
        dp[k][j] = max(dp[k - 1][ptr], arr[j] - arr[ptr]);
      }
    }
    for (long long int k = 2; k <= n; k++) {
      for (long long int j = i; j <= n; j++)
        dp[k][j] = min(dp[k - 1][j], dp[k][j]);
    }
    for (obj x : query[i]) {
      long long int s = x.s, f = x.f, c = x.c, r = x.r;
      res = max(res, dp[r][f] * c);
    }
  }
  cout << res << en;
  return 0;
}