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>
int ss[250000], ff[250000], cc[250000], rr[250000], ii[250000];
int compare(const void *a, const void *b) {
  int i = *(int *)a;
  int j = *(int *)b;
  return rr[i] - rr[j];
}
int main() {
  static int aa[400];
  typedef int ian[400];
  static ian dp_[400], dq_[400], *dp = dp_, *dq = dq_, *tmp;
  int n, m, h, h_, i, r, s, f;
  long long v;
  scanf("%d%d", &n, &m);
  for (i = 0; i < n; i++) scanf("%d", &aa[i]);
  for (h = 0; h < m; h++) {
    scanf("%d%d%d%d", &ss[h], &ff[h], &cc[h], &rr[h]);
    ss[h]--;
    ff[h]--;
    ii[h] = h;
  }
  qsort(ii, m, sizeof *ii, compare);
  for (s = 0; s < n; s++)
    for (f = s + 1; f < n; f++) dp[s][f] = aa[f] - aa[s];
  v = 0;
  for (r = 0, h_ = 0; r <= n; r++) {
    while (h_ < m && rr[h = ii[h_]] == r) {
      long long x;
      if (v < (x = (long long)dp[ss[h]][ff[h]] * cc[h])) v = x;
      h_++;
    }
    for (s = 0; s < n; s++) {
      i = s;
      for (f = s + 1; f < n; f++) {
        while (i < f && dp[s][i] < aa[f] - aa[i]) i++;
        dq[s][f] = aa[f] - aa[i - 1];
        if (dq[s][f] > dp[s][i]) dq[s][f] = dp[s][i];
      }
    }
    tmp = dp;
    dp = dq;
    dq = tmp;
  }
  printf("%lld\n", v);
  return 0;
}