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;
const long long MAX = 3e5;
long long a[MAX];
struct Truck {
  long long s, f, c, r;
} T[MAX];
bool isPossible(long long pos, long long V) {
  long long tank = V;
  long long r = T[pos].r;
  for (long long i = T[pos].s; i < T[pos].f; i++) {
    if (V - T[pos].c * (a[i + 1] - a[i]) < 0) {
      return false;
    }
    if (tank - T[pos].c * (a[i + 1] - a[i]) < 0) {
      if (r > 0) {
        r--;
        tank = V;
      } else {
        return false;
      }
    }
    tank -= T[pos].c * (a[i + 1] - a[i]);
  }
  return true;
}
long long binarySearch(long long pos, long long ini, long long fin) {
  if (!isPossible(pos, fin)) return -1;
  if (isPossible(pos, ini)) return ini;
  while (fin - ini > 1) {
    long long med = (ini + fin) / 2;
    if (isPossible(pos, med)) {
      fin = med;
    } else {
      ini = med;
    }
  }
  return fin;
}
long long n, m;
vector<long long> indices;
long long minimumFill() {
  long long ini = 0;
  long long fin = 5e18;
  long long x = binarySearch(indices[0], ini, fin);
  for (long long i = 1; i < m; i++) {
    if (isPossible(indices[i], x)) {
      continue;
    }
    x = binarySearch(indices[i], x, fin);
  }
  return x;
}
void doit() {
  cin >> n >> m;
  for (long long i = 0; i < (long long)n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < (long long)m; i++) {
    cin >> T[i].s >> T[i].f >> T[i].c >> T[i].r;
    T[i].s--;
    T[i].f--;
    indices.push_back(i);
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  shuffle(indices.begin(), indices.end(), rng);
  cout << minimumFill() << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  doit();
  return 0;
}