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;
template <typename T1, typename T2>
std::istream& operator>>(std::istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
std::ostream& operator<<(std::ostream& out, pair<T1, T2>& a) {
  out << a.first << " " << a.second;
  return out;
}
const long long N = 0;
long long a[405];
vector<pair<pair<long long, long long>, pair<long long, long long>>> v;
bool check(long long s, long long f, long long c, long long r, long long v) {
  long long cnt = 0;
  s--, f--;
  long long cur = v;
  for (long long i = s; i < f; ++i) {
    long long diff = a[i + 1] - a[i];
    if (diff * c > v) return 0;
    if (cur >= diff * c) {
      cur -= diff * c;
    } else {
      cnt++;
      cur = v - diff * c;
    }
  }
  return cnt <= r;
}
long long check(pair<pair<long long, long long>, pair<long long, long long>> p,
                long long v) {
  return check(p.first.first, p.first.second, p.second.first, p.second.second,
               v);
}
void solve() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  v.resize(m);
  for (long long i = 0; i < m; i++) {
    cin >> v[i];
  }
  shuffle((v).begin(), (v).end(), rng);
  long long cm = 0;
  for (long long i = 0; i < m; i++) {
    if (check(v[i], cm + 1)) continue;
    long long L = cm, U = 1e18 + 4;
    while (L <= U) {
      long long mid = (L + U) / 2;
      if (check(v[i], mid)) {
        U = mid - 1;
      } else {
        L = mid + 1;
        cm = max(cm, mid);
      }
    }
  }
  cout << cm + 1;
}
signed main() {
  ios_base::sync_with_stdio(false);
  long long t = 1;
  while (t--) solve();
  return 0;
}