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 <class A>
void rd(vector<A>& v);
template <class T>
void rd(T& x) {
  cin >> x;
}
template <class H, class... T>
void rd(H& h, T&... t) {
  rd(h);
  rd(t...);
}
template <class A>
void rd(vector<A>& x) {
  for (auto& a : x) rd(a);
}
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
void __p(T a) {
  cout << a;
}
template <typename T, typename F>
void __p(pair<T, F> a) {
  cout << "{";
  __p(a.first);
  cout << ",";
  __p(a.second);
  cout << "}\n";
}
template <typename T>
void __p(std::vector<T> a) {
  cout << "{";
  for (auto it = a.begin(); it < a.end(); it++)
    __p(*it), cout << ",}\n"[it + 1 == a.end()];
}
template <typename T, typename... Arg>
void __p(T a1, Arg... a) {
  __p(a1);
  __p(a...);
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : ";
  __p(arg1);
  cout << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  int64_t bracket = 0, i = 0;
  for (;; i++)
    if (names[i] == ',' && bracket == 0)
      break;
    else if (names[i] == '(')
      bracket++;
    else if (names[i] == ')')
      bracket--;
  const char* comma = names + i;
  cout.write(names, comma - names) << " : ";
  __p(arg1);
  cout << " | ";
  __f(comma + 1, args...);
}
void setIO(string s = "") {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin.exceptions(cin.failbit);
  if (((int64_t)(s).size())) {
    freopen((s + ".in").c_str(), "r", stdin);
    freopen((s + ".out").c_str(), "w", stdout);
  }
}
const int64_t MAXM = 250000, MAXN = 400;
int64_t n, m;
int64_t a[MAXN], s[MAXM], e[MAXM], c[MAXM], r[MAXM];
bool chk(int64_t i, int64_t V) {
  int64_t now = V, refill = r[i], cost = c[i];
  for (int64_t j = s[i]; j < e[i]; ++j) {
    int64_t dis = a[j + 1] - a[j];
    int64_t total_cost = dis * cost;
    if (now >= total_cost) {
      now -= total_cost;
    } else {
      if (refill == 0) {
        return false;
      }
      refill--;
      now = V;
      if (total_cost <= now) {
        now -= total_cost;
      } else {
        return false;
      }
    }
  }
  return true;
}
signed main() {
  setIO();
  int64_t n, m;
  rd(n, m);
  for (int64_t i = int64_t(0); i < int64_t(n); ++i) rd(a[i]);
  vector<int64_t> v(m);
  for (int64_t i = int64_t(0); i < int64_t(m); ++i) v[i] = i;
  random_shuffle((v).begin(), (v).end());
  for (auto i : v) {
    rd(s[i], e[i], c[i], r[i]);
    s[i]--, e[i]--;
  }
  int64_t now = 1;
  for (int64_t i = int64_t(0); i < int64_t(m); ++i) {
    if (chk(i, now)) continue;
    int64_t lo = now, hi = 1e18;
    while (hi - lo > 1) {
      int64_t mid = (hi + lo) / 2;
      chk(i, mid) ? hi = mid : lo = mid;
    }
    now = hi;
  }
  cout << now;
}