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 long long MOD = 1e9 + 7;
const long long dx[4] = {-1, 1, 0, 0};
const long long dy[4] = {0, 0, -1, 1};
const long long MAX = 3e5;
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long n, m, k, Q, l, x, y, ans;
string second;
long long a[MAX];
typedef struct Vehicule {
  int from;
  int to;
  long long cons;
  int refuel;
};
Vehicule v[MAX];
bool check(long long tank, int j) {
  int refsofar = 0;
  long long level = tank;
  for (int i = v[j].from; i < v[j].to; ++i) {
    long long dist = a[i + 1] - a[i];
    long long fuelneeded = dist * v[j].cons;
    if (fuelneeded > level) {
      if (refsofar == v[j].refuel || fuelneeded > tank) return false;
      refsofar++;
      level = tank;
    }
    level -= fuelneeded;
  }
  return true;
}
long long new_tank(int i) {
  long long lo = 1, hi = 1e18;
  long long mid;
  long long ans = 1e18;
  while (lo <= hi) {
    mid = (lo + hi) >> 1;
    if (check(mid, i)) {
      ans = mid;
      hi = mid - 1;
    } else
      lo = mid + 1;
  }
  return ans;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) {
    cin >> v[i].from >> v[i].to >> v[i].cons >> v[i].refuel;
    v[i].from--;
    v[i].to--;
  }
  random_shuffle(v, v + m);
  ans = 1;
  for (int i = 0; i < m; ++i) {
    if (check(ans, i)) continue;
    ans = new_tank(i);
  }
  cout << ans << '\n';
  return 0;
}