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 N = 405, M = 250005;
int n, m, p[N], a[M], b[M], c[M], d[M];
vector<int> id[N];
long long f[N][N][2], ans;
int read() {
  int x = 0, f = 1;
  char s;
  while ((s = getchar()) > '9' || s < '0')
    if (s == '-') f = -1;
  while (s >= '0' && s <= '9')
    x = (x << 1) + (x << 3) + (s ^ 48), s = getchar();
  return x * f;
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; ++i) p[i] = read();
  for (int i = 1; i <= m; ++i) {
    a[i] = read(), b[i] = read(), c[i] = read(), d[i] = read() + 1;
    d[i] = min(d[i], b[i] - a[i]);
    id[d[i]].push_back(i);
  }
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j) f[i][j][0] = p[j] - p[i];
  for (int k = 1; k < n; ++k) {
    for (int i = 0; i < id[k].size(); ++i)
      ans = max(ans, f[a[id[k][i]]][b[id[k][i]]][k + 1 & 1] * c[id[k][i]]);
    for (int i = 1; i <= n; ++i) {
      f[i][i][k & 1] = 0;
      int pos = i;
      for (int j = i + 1; j <= n; ++j) {
        while (f[i][pos][k + 1 & 1] < p[j] - p[pos]) ++pos;
        f[i][j][k & 1] = min(f[i][pos][k + 1 & 1], 0ll + p[j] - p[pos - 1]);
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}