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 = 4e2 + 10;
const int inf = 1e9 + 10;
int n;
int a[N];
vector<int> f[N][N];
bool minimize(int &x, int y) {
  if (x > y)
    x = y;
  else
    return 0;
  return 1;
}
int get(int L, int R, int use) {
  use = min(use, R - L);
  return f[L][R][use];
}
void DP(int L, int R, int uL, int uR, int optL, int optR) {
  int uM = (uL + uR) / 2;
  int &ans = f[L][R][uM] = inf;
  int opt = -1;
  for (int i = optL; i <= optR; i++) {
    if (minimize(ans, max(get(L, i, uM - 1), a[R] - a[i]))) opt = i;
  }
  if (uM > uL) {
    DP(L, R, uL, uM - 1, optL, opt);
  }
  if (uM < uR) {
    DP(L, R, uM + 1, uR, opt, optR);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int test;
  cin >> n >> test;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int L = n; L >= 1; L--) {
    f[L][L] = vector<int>(1, 0);
    for (int R = L + 1; R <= n; R++) {
      f[L][R] = vector<int>(R - L + 1, 0);
      f[L][R][0] = a[R] - a[L];
      DP(L, R, 1, R - L, L, R - 1);
    }
  }
  long long ans = 0;
  while (test--) {
    int L, R, cost, use;
    cin >> L >> R >> cost >> use;
    ans = max(ans, 1LL * get(L, R, use) * cost);
  }
  cout << ans;
}