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 inf = 2e9;
int n, m, a[403];
int memo[403][403][403];
long long ans;
int rec(int l, int r, int k) {
  if (l == r) {
    return 0;
  }
  if (k == 0) {
    return (a[r] - a[l]);
  }
  if (memo[l][r][k] != inf) {
    return memo[l][r][k];
  }
  int lo = l, hi = r - 1;
  while (lo < hi) {
    int mid = (lo + hi) / 2;
    if ((rec(l, mid, k - 1) - (a[r] - a[mid])) < 0) {
      lo = mid + 1;
    } else {
      hi = mid;
    }
  }
  memo[l][r][k] = max(rec(l, lo, k - 1), a[r] - a[lo]);
  if (lo > l) {
    memo[l][r][k] =
        min(memo[l][r][k], max(rec(l, lo - 1, k - 1), a[r] - a[lo - 1]));
  }
  return memo[l][r][k];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < 403; ++i) {
    for (int j = 0; j < 403; ++j) {
      for (int k = 0; k < 403; ++k) {
        memo[i][j][k] = inf;
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    int s, f, c, r;
    cin >> s >> f >> c >> r;
    --s, --f;
    ans = max(ans, 1LL * c * rec(s, f, r));
  }
  cout << ans;
}