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 N = 250000 + 77, L = 400 + 77;
long long n, m, pos[L][L], a[L], St[N], En[N], C[N], R[N];
long long dp[L][L], A;
vector<int> V[L];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= m; ++i) {
    cin >> St[i] >> En[i] >> C[i] >> R[i];
    V[St[i]].push_back(i);
  }
  for (int le = 1; le <= n; ++le) {
    memset(dp, 0, sizeof(dp));
    memset(pos, 0, sizeof(pos));
    for (int ri = le; ri <= n; ++ri) dp[ri][0] = a[ri] - a[le], pos[ri][0] = le;
    for (int k = 1; k <= n; ++k) dp[le][k] = 0, pos[le][k] = le;
    for (int ri = le + 1; ri <= n; ++ri)
      for (int k = 1; k <= n; ++k) {
        int p = pos[ri - 1][k];
        dp[ri][k] = max(dp[p][k - 1], a[ri] - a[p]);
        pos[ri][k] = p;
        for (int t = p + 1; t < ri; ++t) {
          if (max(dp[t][k - 1], a[ri] - a[t]) <= dp[ri][k]) {
            dp[ri][k] = max(dp[t][k - 1], a[ri] - a[t]);
            pos[ri][k] = t;
          } else {
            break;
          }
        }
      }
    for (auto x : V[le]) {
      A = max(A, dp[En[x]][R[x]] * 1ll * C[x]);
    }
  }
  cout << A;
  return 0;
}