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 long double PI = 3.14159265358979323846;
const long long N = 250005;
const long long M = 401;
long long pos[M];
long long s[N], f[N], c[N], r[N];
long long n, m;
void pre() {}
bool chk(long long ind, long long mnb) {
  long long rfl = 0, rem = mnb;
  bool imp = 0;
  for (long long i = s[ind] + 1; i <= (f[ind]); ++i) {
    long long gp = (pos[i] - pos[i - 1]) * c[ind];
    if (gp > mnb) {
      imp = 1;
      break;
    }
    if (gp > rem) {
      rem = mnb;
      rfl++;
    }
    rem -= gp;
  }
  if (rfl <= r[ind] and !imp) return true;
  return false;
}
long long getA(long long ind, long long mnb) {
  if (chk(ind, mnb)) return mnb;
  long long l = mnb + 1, h = 1e18, ans = -1;
  while (h >= l) {
    long long mid = l + (h - l) / 2;
    if (chk(ind, mid)) {
      ans = mid;
      h = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tt = 1;
  for (long long ttt = 1; ttt <= (tt); ++ttt) {
    cin >> n >> m;
    for (long long i = 0; i < (n); ++i) cin >> pos[i];
    for (long long i = 0; i < (m); ++i) {
      cin >> s[i] >> f[i] >> c[i] >> r[i];
      s[i]--;
      f[i]--;
    }
    vector<long long> ind(m);
    iota((ind).begin(), (ind).end(), 0);
    random_shuffle((ind).begin(), (ind).end());
    long long ans = 0;
    for (long long i = 0; i < (m); ++i) {
      ans = getA(ind[i], ans);
    }
    cout << ans << '\n';
  }
}