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 eps = 1e-13;
const long double PI = acos(-1);
const int INF = (int)1e9;
const long long INFF = (long long)1e18;
const int mod = 998244353;
const int MXN = (int)4e2 + 7;
inline long long add(long long v1, long long v2) {
  v1 += v2;
  if (v1 >= mod) v1 -= mod;
  if (v1 < 0) v1 += mod;
  return v1;
};
int a[MXN];
vector<pair<int, int> > qy[MXN][MXN];
int dp[MXN][MXN][2];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  long long ans = 0;
  while (m--) {
    int l, r, c, t;
    scanf("%d %d %d %d", &l, &r, &c, &t);
    t = min(t, r - l - 1);
    qy[t][l].push_back(make_pair(r, c));
  }
  int p1 = 0, p2 = 1;
  for (int i = 1; i < n; i++) {
    for (int j = i + 1; j <= n; j++) dp[i][j][0] = a[j] - a[i];
    for (pair<int, int> x : qy[0][i]) {
      ans = max(ans, dp[i][x.first][0] * 1ll * x.second);
    }
  }
  for (int i = 1; i < n; i++, swap(p1, p2)) {
    for (int l = 1; l < n; l++) {
      int p = l + i;
      for (int r = l + i + 1; r <= n; r++) {
        while (max(dp[l][p][p1], a[r] - a[p]) >=
               max(dp[l][p + 1][p1], a[r] - a[p + 1]))
          p++;
        dp[l][r][p2] = max(dp[l][p][p1], a[r] - a[p]);
      }
      for (pair<int, int> x : qy[i][l]) {
        ans = max(ans, dp[l][x.first][p2] * 1ll * x.second);
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}