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;
long long nxt() {
  long long x;
  cin >> x;
  return x;
}
void setIO(string s) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
void yes() { cout << "YES\n"; }
void no() { cout << "NO\n"; }
const long long mod = 1e9 + 7;
const long long INF = 2e18;
struct truck {
  long long s, f, c, r;
  truck(int _s, int _f, int _c, int _r) : s(_s), f(_f), c(_c), r(_r) {}
  truck() {}
};
bool possible(long long V, truck &t, vector<long long> &a) {
  42;
  42;
  int it = t.s;
  int cnt = 0;
  bool ans = 1;
  while (it < t.f) {
    cnt++;
    if (cnt > t.r + 1) return 0;
    int it1 = it;
    long long curV = V;
    42;
    while (it1 + 1 <= t.f && (a[it1 + 1] - a[it1]) * t.c <= curV) {
      curV -= (a[it1 + 1] - a[it1]) * t.c;
      it1++;
      42;
    }
    if (it == it1)
      ans = 0;
    else
      it = it1;
  }
  42;
  return ans;
}
void solve() {
  unsigned seed1 = std::chrono::system_clock::now().time_since_epoch().count();
  std::mt19937 rng(seed1);
  int n, m;
  cin >> n >> m;
  vector<long long> a(n);
  for (int i = (int)(0); i < (int)(n); i++) cin >> a[i];
  vector<truck> v(m);
  for (int i = (int)(0); i < (int)(m); i++) {
    long long s, f, c, r;
    cin >> s >> f >> c >> r;
    s--;
    f--;
    v[i] = truck(s, f, c, r);
  }
  shuffle(v.begin(), v.end(), rng);
  42;
  long long curV = 0;
  for (int i = (int)(0); i < (int)(m); i++) {
    42;
    if (possible(curV, v[i], a)) continue;
    long long lo = curV + 1, hi = INF;
    while (lo < hi) {
      42;
      long long mid = (lo + hi) / 2;
      if (possible(mid, v[i], a)) {
        hi = mid;
      } else {
        lo = mid + 1;
      }
    }
    curV = max(curV, lo);
  }
  cout << curV << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(10);
  int T = 1;
  for (int tc = 1; tc <= T; tc++) {
    solve();
  }
}