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 maxn_m = 5e5 + 10, maxn_n = 4e2 + 10;
int arr[maxn_n], n, m;
long long dp[maxn_n][maxn_n];
vector<pair<pair<int, int>, pair<int, int> > > vec;
void input() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < m; i++) {
    int s, f, c, r;
    cin >> s >> f >> c >> r, s--, f--;
    vec.push_back({{s, f}, {c, r}});
  }
}
bool check(int j, long long v) {
  int s = vec[j].first.first, f = vec[j].first.second, c = vec[j].second.first,
      r = vec[j].second.second, t = 0;
  long long sum = v;
  for (int i = s; i < f; i++)
    if ((long long)c * (arr[i + 1] - arr[i]) > v)
      return false;
    else if ((long long)c * (arr[i + 1] - arr[i]) > sum)
      sum = v - (long long)c * (arr[i + 1] - arr[i]), t++;
    else
      sum -= (long long)c * (arr[i + 1] - arr[i]);
  return (t <= r);
}
long long dpp(int x) {
  int s = vec[x].first.first, f = vec[x].first.second, c = vec[x].second.first,
      r = vec[x].second.second, t = 0;
  for (int i = 0; i < n + 1; i++)
    for (int j = 0; j < n; j++) dp[i][j] = 0;
  for (int i = s; i < f + 1; i++) dp[0][i] = (long long)c * (arr[i] - arr[s]);
  for (int i = 1; i < r + 1; i++) {
    int ptr = s;
    for (int j = s; j < f + 1; j++) {
      long long ansptr =
          max(dp[i - 1][ptr], (long long)c * (arr[j] - arr[ptr]));
      for (int k = ptr; k < f + 1; k++) {
        long long ans = max(dp[i - 1][k], (long long)c * (arr[j] - arr[k]));
        if (ans < ansptr) ansptr = ans, ptr = k;
      }
      dp[i][j] = ansptr;
    }
  }
  return dp[r][f];
}
long long ans(int x, long long l) {
  long long r = 1e18 + 10;
  while (r - l > 1) {
    long long mid = (l + r) / 2;
    if (check(x, mid))
      r = mid;
    else
      l = mid;
  }
  return r;
}
int main() {
  input();
  srand(time(0));
  random_shuffle(vec.begin(), vec.end());
  long long v = 0;
  for (int i = 0; i < m; i++)
    if (!check(i, v)) v = dpp(i);
  cout << v << endl;
}