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 = 407;
const int maxb = 29;
const long long ool = 1e18 + 7;
class car {
 public:
  long long s;
  long long t;
  long long r;
  long long c;
  car(long long ss = 1, long long tt = 1, long long cc = 1, long long rr = 1) {
    s = ss;
    t = tt;
    r = rr;
    c = cc;
  }
};
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<long long> pos(n + 1);
  vector<vector<vector<long long> > > dp(
      2, vector<vector<long long> >(n + 1, vector<long long>(n + 1)));
  vector<vector<car> > ck(n + 1);
  for (int i = (1); i <= (n); ++i) cin >> pos[i];
  for (int i = (0); i < (m); ++i) {
    int s, t, r, c;
    cin >> s >> t >> c >> r;
    int kk = min(t - s - 1, r) + 1;
    ck[kk].emplace_back(car(s, t, c, kk));
  }
  int t = 0;
  long long ans = 0;
  for (int i = (1); i <= (n); ++i)
    for (int j = (i); j <= (n); ++j)
      dp[t][i][j] = (i == j ? 0 : pos[j] - pos[i]);
  for (int k = (1); k < (n); ++k) {
    for (auto x : ck[k]) ans = max(dp[t][x.s][x.t] * x.c, ans);
    if (k == n - 1) break;
    for (int i = (1); i <= (n); ++i)
      for (int j = (i); j <= (n); ++j) dp[t ^ 1][i][j] = ool;
    for (int i = (1); i <= (n); ++i) {
      dp[t ^ 1][i][i] = 0;
      int x = i;
      for (int j = (i + 1); j <= (n); ++j) {
        while (dp[t][i][x] < pos[j] - pos[x]) x++;
        dp[t ^ 1][i][j] = min(dp[t][i][x], pos[j] - pos[x - 1]);
      }
    }
    t ^= 1;
  }
  cout << ans << endl;
  return 0;
}