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. 
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const int MAXN = 401;
const int INF = 1e9 + 5;
int city[MAXN], dp[MAXN][MAXN][MAXN];
int main() {
  for (int i = 0; i < int(MAXN); i++)
    for (int j = 0; j < int(MAXN); j++)
      for (int k = 0; k < int(MAXN); k++) dp[i][j][k] = INF;
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < int(n); i++) scanf("%d", &city[i]);
  for (int i = 0; i < int(n); i++)
    for (int j = int(i); j < int(n); j++) dp[0][i][j] = city[j] - city[i];
  for (int k = int(1); k < int(n + 1); k++)
    for (int i = 0; i < int(n); i++) {
      int ind = i;
      for (int j = int(i); j < int(n); j++) {
        dp[k][i][j] = max(dp[k - 1][i][ind], city[j] - city[ind]);
        while (ind + 1 <= j && max(dp[k - 1][i][ind + 1],
                                   city[j] - city[ind + 1]) < dp[k][i][j]) {
          ind++;
          dp[k][i][j] =
              min(dp[k][i][j], max(dp[k - 1][i][ind], city[j] - city[ind]));
        }
      }
    }
  long long rta = 0;
  for (int i = 0; i < int(m); i++) {
    int s, f, c, r;
    scanf("%d %d %d %d", &s, &f, &c, &r);
    s--;
    f--;
    int val = dp[r][s][f];
    rta = max(rta, (long long)val * c);
  }
  printf("%lld", rta);
  return 0;
}
