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 M = 2.5E5 + 5;
int n, m;
int a[401], b[401], s[M], f[M], c[M], r[M], r1[M], i, in, dis;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long val;
bool check() {
  dis = 0;
  r1[in] = r[in];
  for (i = s[in]; i < f[in]; i++) {
    if (1ll * dis * c[in] + 1ll * b[i] * c[in] > val) {
      r1[in]--;
      if (r1[in] < 0 || !dis) return 0;
      dis = 0;
      i--;
    } else
      dis += b[i];
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) b[i] = a[i + 1] - a[i];
  long long MAX = 0;
  for (i = 0; i < m; i++) {
    cin >> s[i] >> f[i] >> c[i] >> r[i];
  }
  vector<int> v(m);
  for (i = 0; i < m; i++) v[i] = i;
  shuffle(v.begin(), v.end(), rng);
  long long l, r1;
  for (int i = 0; i < m; i++) {
    in = v[i];
    val = MAX;
    if (check()) continue;
    l = MAX + 1;
    MAX = 1e18;
    while (l != MAX) {
      val = (l + MAX) / 2;
      if (check())
        MAX = val;
      else
        l = val + 1;
    }
  }
  cout << MAX;
}