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 long MOD = 1e9 + 7;
const long long dx[4] = {-1, 1, 0, 0};
const long long dy[4] = {0, 0, -1, 1};
const long long MAX = 3e5;
const long long oo = 1e16;
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long n, m, k, Q, l, x, y, ans;
string second;
long long a[MAX];
typedef struct Vehicule {
  int from;
  int to;
  long long cons;
  int refuel;
};
Vehicule v[MAX];
long long dp[500][500];
vector<vector<int> > st;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  st.clear();
  st.resize(n + 1);
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) {
    cin >> v[i].from >> v[i].to >> v[i].cons >> v[i].refuel;
    v[i].from--;
    v[i].to--;
    st[v[i].from].push_back(i);
  }
  ans = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) dp[i][j] = oo;
  for (int l = 0; l < n; ++l) {
    for (int r = l; r < n; ++r) dp[r][1] = a[r] - a[l];
    for (int r = l + 1; r < n; ++r)
      for (int k = 2; k < n + 5; ++k) {
        long long best = oo;
        int j = r;
        while (j >= l && max(dp[j][k - 1], a[r] - a[j]) <= best)
          best = max(dp[j][k - 1], a[r] - a[j]), --j;
        dp[r][k] = best;
      }
    for (auto u : st[l])
      ans = max(ans, dp[v[u].to][v[u].refuel + 1] * 1LL * v[u].cons);
  }
  cout << ans << '\n';
  return 0;
}