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 int INF = INT_MAX / 2;
class Solution {
 public:
  long long minimize(int n, int m, vector<int>& pos, vector<int>& from,
                     vector<int>& to, vector<int>& cost, vector<int>& limit) {
    vector<vector<int>> dp(n, vector<int>(n, INF));
    for (int i = 0; i < n; ++i) {
      dp[i][i] = 0;
    }
    vector<int> indices(m, 0);
    iota(indices.begin(), indices.end(), 0);
    auto cmp = [&](int l, int r) { return limit[l] < limit[r]; };
    sort(indices.begin(), indices.end(), cmp);
    int L = limit[indices.back()];
    long long res = 0;
    for (int l = 1, t = 0; l <= L + 1; ++l) {
      for (int i = n - 1; i >= 0; --i) {
        int r = i - 1;
        for (int j = i - 1; j >= 0; --j) {
          for (int k = r; k >= j; --k) {
            dp[j][i] = min(dp[j][i], max(dp[j][k], pos[i] - pos[k]));
            if (dp[j][k] >= pos[i] - pos[k]) {
              r = k;
            } else {
              break;
            }
          }
        }
      }
      while (t < m && limit[indices[t]] + 1 == l) {
        res = max(
            res, 1LL * cost[indices[t]] * dp[from[indices[t]]][to[indices[t]]]);
        ++t;
      }
    }
    return res;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution sol;
  int n, m;
  cin >> n >> m;
  vector<int> pos(n);
  for (int i = 0; i < n; ++i) {
    cin >> pos[i];
  }
  vector<int> from(m);
  vector<int> to(m);
  vector<int> cost(m);
  vector<int> limit(m);
  for (int i = 0; i < m; ++i) {
    cin >> from[i] >> to[i] >> cost[i] >> limit[i];
    --from[i];
    --to[i];
  }
  cout << sol.minimize(n, m, pos, from, to, cost, limit) << endl;
  return 0;
}