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;
constexpr long long INF = 1e18;
int main() {
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  vector<array<int, 4>> trucks(m);
  for (auto& truck : trucks) {
    cin >> truck[0] >> truck[1] >> truck[2] >> truck[3];
    --truck[0];
    --truck[1];
    if (truck[3] == n) {
      --truck[3];
    }
  }
  sort(trucks.begin(), trucks.end());
  long long ans = 0;
  int curr = 0;
  for (int left = 0; left < n; ++left) {
    vector<vector<int>> dp(1 + n, vector<int>(n, 0));
    for (int right = left; right < n; ++right) {
      dp[1][right] = a[right] - a[left];
    }
    for (int k = 2; k <= n; ++k) {
      int opt = left;
      dp[k][left] = 0;
      for (int right = left + 1; right < n; ++right) {
        while (opt + 1 < right &&
               max(dp[k - 1][opt], a[right] - a[opt]) >
                   max(dp[k - 1][opt + 1], a[right] - a[opt + 1])) {
          ++opt;
        }
        dp[k][right] = max(dp[k - 1][opt], a[right] - a[opt]);
      }
    }
    while (curr < m && trucks[curr][0] <= left) {
      if (trucks[curr][0] == left) {
        ans = max(ans, 1LL * dp[trucks[curr][3] + 1][trucks[curr][1]] *
                           trucks[curr][2]);
      }
      ++curr;
    }
  }
  cout << ans << endl;
}