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;
vector<string> vec_splitter(string s) {
  for (char& c : s) c = c == ',' ? ' ' : c;
  stringstream ss;
  ss << s;
  vector<string> res;
  for (string z; ss >> z; res.push_back(z))
    ;
  return res;
}
void debug_out(vector<string> args, int idx) { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, Head H, Tail... T) {
  if (idx > 0) cerr << ", ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, T...);
}
const long long N = 405, M = 3e5 + 5, D = 20000;
long long A[N], second[M], first[M], C[M], R[M], n, m, st = 1;
bool doit(long long idx, long long x) {
  long long city = second[idx], refuel = R[idx], currFuel = x;
  while (city != first[idx]) {
    long long fuelReq = (A[city + 1] - A[city]) * C[idx];
    if (fuelReq <= currFuel) {
      currFuel -= fuelReq;
      ++city;
      continue;
    }
    if (!refuel) return 0;
    refuel--;
    currFuel = x;
    if (fuelReq <= currFuel) {
      currFuel -= fuelReq;
      ++city;
      continue;
    }
    return 0;
  }
  return 1;
}
bool solve(long long x) {
  for (long long i = st; i <= m; ++i) {
    if (!doit(i, x)) {
      st = i;
      return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; ++i) scanf("%lld", &A[i]);
  for (long long i = 1; i <= m; ++i)
    scanf("%lld%lld", &second[i], &first[i]), scanf("%lld%lld", &C[i], &R[i]);
  long long lo = 1, hi = 1e18, ans = 0;
  while (lo <= hi) {
    long long mid = lo + (hi - lo + 1) / D;
    if (solve(mid))
      ans = mid, hi = mid - 1;
    else
      lo = mid + 1;
  }
  printf("%lld\n", ans);
  return 0;
}