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;
typedef struct {
  int ss;
  int ff;
  int rr;
  int cc;
} inp;
inp in[250005];
int a[405];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  srand(time(NULL));
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long long ans = 0;
  for (int i = 1; i <= m; i++) {
    cin >> in[i].ss >> in[i].ff >> in[i].cc >> in[i].rr;
  }
  reverse(in + 1, in + 1 + m);
  int ms = 0, mf = 0, mc = 0, mr = 0;
  for (int i = 1; i <= m; i++) {
    int s = in[i].ss, f = in[i].ff, c = in[i].cc, r = in[i].rr;
    int cur = s;
    if (a[mf] - a[ms] >= a[f] - a[s] && c <= mc && r >= mr) continue;
    for (int j = 0; j <= r; j++) {
      long long LL = s, RR = n;
      while (LL != RR) {
        long long midd = (LL + RR + 1) / 2;
        if (1LL * (a[midd] - a[cur]) * c <= ans)
          LL = midd;
        else
          RR = midd - 1;
      }
      cur = LL;
      if (cur >= f) break;
    }
    if (cur >= f) continue;
    long long L = cur, R = 1e18;
    while (L != R) {
      long long mid = (L + R) / 2;
      cur = s;
      for (int j = 0; j <= r; j++) {
        long long LL = s, RR = n;
        while (LL != RR) {
          long long midd = (LL + RR + 1) / 2;
          if (1LL * (a[midd] - a[cur]) * c <= mid)
            LL = midd;
          else
            RR = midd - 1;
        }
        cur = LL;
        if (cur >= f) break;
      }
      if (cur >= f)
        R = mid;
      else
        L = mid + 1;
    }
    ans = L;
    ms = s;
    mf = f;
    mc = c;
    mr = r;
  }
  cout << ans;
}