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 mod = 998244353;
const int inf = 1e9 + 9;
const long long int MOD = 1e9 + 696969;
const long long INF = 4LL * 1000000LL * 1000000LL * 1000000LL;
template <class T>
int size(T &&x) {
  return int(x.size());
}
template <class A, class B>
ostream &operator<<(ostream &out, const pair<A, B> &p) {
  return out << '(' << p.first << ", " << p.second << ')';
}
template <class T>
auto operator<<(ostream &out, T &&x) -> decltype(x.begin(), out) {
  out << '{';
  for (auto it = x.begin(); it != x.end(); ++it)
    out << *it << (it == prev(x.end()) ? "" : ", ");
  return out << '}';
}
void dump() {}
template <class T, class... Args>
void dump(T &&x, Args... args) {
  cerr << x << ";  ";
  dump(args...);
}
mt19937_64 rng(0);
int random(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
inline void addmod(int &x, int val) {
  x += val;
  while (x >= mod) x -= mod;
}
const int maxn = 250100;
int arr[maxn];
int pocz[maxn], kon[maxn], dozw[maxn], n, m;
long long cons[maxn];
int perm[maxn];
bool check(int kt, long long v) {
  int a = pocz[kt], b = kon[kt];
  int res = 0;
  long long fuel = v;
  for (int i = a; i < b; ++i) {
    long long will_take = cons[kt] * (arr[i + 1] - arr[i]);
    if (will_take > v) return false;
    if (will_take > fuel) {
      ++res;
      fuel = v - will_take;
    } else
      fuel -= will_take;
  }
  return res <= dozw[kt];
}
int main() {
  srand(time(0));
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  };
  cin >> n >> m;
  for (int i = (1); i <= (n); ++i) cin >> arr[i];
  for (int i = (1); i <= (m); ++i) {
    cin >> pocz[i] >> kon[i] >> cons[i] >> dozw[i];
    perm[i] = i;
  }
  long long V = 1;
  long long MX = (long long)1e18;
  MX += 3;
  random_shuffle(perm + 1, perm + m + 1);
  for (int i = 1; i <= m; ++i) {
    if (!check(perm[i], V)) {
      long long x = V + 1, y = MX;
      while (x < y) {
        long long sr = (x + y) / 2LL;
        if (!check(perm[i], sr))
          x = ++sr;
        else
          y = sr;
      }
      V = x;
    }
  }
  cout << V;
}