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 Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long maxn = (long long)1e6 + 6;
const long double EPS = 1e-9;
const long long INF = (long long)1e18 + 18;
const long long mod = (long long)998244353;
long long n, m;
long long s[maxn], f[maxn], c[maxn], r[maxn];
vector<long long> a;
bool works(long long i, long long V) {
  long long cur = V, refuel = 0;
  for (long long it = s[i] + 1; it < f[i]; it++) {
    long long d = a[it] - a[it - 1];
    long long next = it + 1;
    if (cur < c[i] * d) {
      return false;
    }
    cur = cur - c[i] * d;
    if (cur < c[i] * (a[next] - a[it])) {
      if (refuel < r[i]) {
        refuel++;
        cur = V;
      } else {
        return false;
      }
    }
  }
  if (cur < c[i] * (a[f[i]] - a[f[i] - 1])) {
    return false;
  }
  return true;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  a.resize(n + 1);
  vector<long long> perm(m);
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= m; i++) {
    cin >> s[i] >> f[i] >> c[i] >> r[i];
    perm[i - 1] = i;
  }
  random_shuffle(perm.begin(), perm.end());
  long long first = 0, last = (long long)1e18;
  for (long long i = 1; i <= m; i++) {
    long long index = perm[i - 1];
    if (works(index, first) == 1) {
      continue;
    }
    long long st = first, en = last;
    while (st < en) {
      long long mid = (st + en) >> 1;
      if (works(index, mid)) {
        en = mid;
      } else {
        st = mid + 1;
      }
    }
    first = st;
  }
  cout << first << '\n';
  return 0;
}