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 MAXN = 400;
const int MAXM = 250000;
const int INF = 0x3f3f3f3f;
int a[MAXN + 5];
int s[MAXM + 5], t[MAXM + 5], c[MAXM + 5], r[MAXM + 5];
vector<int> id[MAXN + 5];
int n, m;
inline void Init() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= m; ++i) {
    scanf("%d %d %d %d", &s[i], &t[i], &c[i], &r[i]), ++r[i];
    r[i] = min(r[i], t[i] - s[i]);
    id[r[i]].push_back(i);
  }
}
int f[MAXN + 5][MAXN + 5][2], now = 0;
inline void Getdp() {
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j) f[i][j][now] = a[j] - a[i];
  long long ans = -INF;
  for (auto i : id[1]) ans = max(ans, 1ll * f[s[i]][t[i]][now] * c[i]);
  for (int k = 2; k <= n; ++k) {
    now ^= 1;
    for (int i = 1; i <= n; ++i)
      for (int j = i, l = i; j <= n; ++j) {
        while (f[i][l][now ^ 1] < a[j] - a[l]) ++l;
        f[i][j][now] = min(f[i][l][now ^ 1], a[j] - a[l - 1]);
      }
    for (auto i : id[k]) ans = max(ans, 1ll * f[s[i]][t[i]][now] * c[i]);
  }
  printf("%lld\n", ans);
}
signed main() {
  Init();
  Getdp();
  return 0;
}