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;
using ll = long long;
template <typename T>
using v = vector<T>;
struct Truck {
  ll s, f, c, r;
};
const ll N = 400;
ll cities[N + 1];
inline ll getnaivecapacity(const Truck &truck) {
  return (cities[truck.f] - cities[truck.s]) * truck.c;
}
inline bool ispos(const Truck &truck, ll capacity) {
  Truck state = truck;
  ll liters = capacity;
  ll curcity = state.s;
  for (; curcity < state.f; curcity++) {
    const ll tripfuel = (cities[curcity + 1] - cities[curcity]) * state.c;
    if (liters >= tripfuel || state.r == 0) {
      liters -= tripfuel;
    } else {
      state.r--;
      liters = capacity - tripfuel;
    }
    if (liters < 0) {
      return false;
    }
  }
  return true;
}
inline ll getoptimalcapacity(const Truck &truck, ll lo, ll hi) {
  while (lo != hi) {
    const ll mid = (lo + hi) >> 1;
    if (ispos(truck, mid)) {
      hi = mid;
    } else {
      lo = mid + 1;
    }
  }
  return lo;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  ll n, m;
  cin >> n >> m;
  for (ll i = 1; i <= n; i++) {
    cin >> cities[i];
  }
  v<Truck> trucks;
  for (ll i = 1; i <= m; i++) {
    ll s, f, c, r;
    cin >> s >> f >> c >> r;
    trucks.push_back(Truck{s, f, c, r});
  }
  shuffle(trucks.begin(), trucks.end(), rng);
  ll lo =
      getoptimalcapacity(trucks.front(), 1, getnaivecapacity(trucks.front()));
  for (ll i = 1; i < (ll)trucks.size(); i++) {
    ll hi = getnaivecapacity(trucks[i]);
    if (hi > lo && !ispos(trucks[i], lo)) {
      lo = getoptimalcapacity(trucks[i], lo + 1, hi);
    }
  }
  cout << lo << "\n";
  return 0;
}