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;
using INT = long long;
const int NN = 402;
int a[NN], dis[NN];
int dp[NN][NN][NN];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      for (int k = 0; k < n; k++) dp[i][j][k] = 1e9;
  for (int i = 1; i < n; i++)
    for (int j = i + 1; j <= n; j++) {
      dp[i][j][0] = a[j] - a[i];
      for (int k = 1; k < j - i; k++) {
        for (int ed = j - 1; ed > i; ed--) {
          if (a[j] - a[ed] > dp[i][ed][k - 1]) {
            if (a[j] - a[ed] < dp[i][ed + 1][k - 1])
              dp[i][j][k] = a[j] - a[ed];
            else
              dp[i][j][k] = dp[i][ed + 1][k - 1];
            break;
          }
          if (ed == i + 1) dp[i][j][k] = a[i + 1] - a[i];
        }
      }
    }
  INT ans = 0;
  while (m--) {
    int st, ed, c, r;
    scanf("%d%d%d%d", &st, &ed, &c, &r);
    r = min(r, ed - st - 1);
    ans = max(ans, 1LL * c * dp[st][ed][r]);
  }
  cout << ans << endl;
  return 0;
}