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 int nax = 3e5 + 10;
struct truck {
  int s, f, c, r;
  truck() {}
  truck(int s, int f, int c, int r) : s(s), f(f), c(c), r(r) {}
  void read() { scanf("%d %d %d %d", &s, &f, &c, &r); }
} x[nax];
mt19937 rng((unsigned long long)(new char()));
const long long lim = 1e18;
int n, m;
int a[nax];
int ind[nax];
long long cur_max;
int get_par(truck &b, long long V) {
  auto [s, f, c, r] = b;
  long long cur_sum = 0;
  int cnt = 0;
  for (int i = s + 1; i <= f; ++i) {
    cur_sum += (a[i] - a[i - 1]);
    if (cur_sum * c > V) {
      ++cnt;
      cur_sum = a[i] - a[i - 1];
    }
    if (cur_sum * c > V) return INT_MAX;
  }
  return cnt;
}
void get_par(truck &b) {
  if (get_par(b, cur_max) <= b.r) return;
  long long l = cur_max, r = lim, ans = -1;
  while (l <= r) {
    long long mid = l + r >> 1;
    if (get_par(b, mid) <= b.r) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cur_max = ans;
  return;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= m; ++i) {
    x[i].read();
  }
  iota(ind + 1, ind + 1 + m, 1);
  shuffle(ind + 1, ind + 1 + m, rng);
  for (int _ = 1; _ <= m; ++_) {
    int i = ind[_];
    get_par(x[i]);
  }
  printf("%lld\n", cur_max);
}