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;
namespace ywy {
inline int get() {
  int n = 0;
  char c;
  while ((c = getchar()) || 23333) {
    if (c >= '0' && c <= '9') break;
    if (c == '-') goto s;
  }
  n = c - '0';
  while ((c = getchar()) || 23333) {
    if (c >= '0' && c <= '9')
      n = n * 10 + c - '0';
    else
      return (n);
  }
s:
  while ((c = getchar()) || 23333) {
    if (c >= '0' && c <= '9')
      n = n * 10 - c + '0';
    else
      return (n);
  }
}
typedef struct _n {
  int s;
  int t;
  int v;
  int lim;
} node;
node ints[300001];
int pos[444];
int mx[444][444];
inline int Tanxin(int x, long long v) {
  if (mx[ints[x].s][ints[x].t] * (long long)ints[x].v > v) return (0);
  int cur = ints[x].s;
  int tot = 0;
  while (cur != ints[x].t && tot <= ints[x].lim) {
    if ((long long)ints[x].v * (pos[ints[x].t] - pos[cur]) <= v) return (1);
    int ans = cur + 1, l = cur + 1, r = ints[x].t - 1;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if ((long long)ints[x].v * (pos[mid + 1] - pos[cur]) > v)
        ans = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    tot++;
    cur = ans;
  }
  if (tot > ints[x].lim) return (0);
  return (1);
}
void ywymain() {
  srand(19260817);
  int n = get(), m = get();
  for (register int i = 1; i <= n; i++) pos[i] = get();
  for (register int i = 1; i <= m; i++) {
    ints[i].s = get();
    ints[i].t = get();
    ints[i].v = get();
    ints[i].lim = get();
  }
  for (register int i = 1; i <= n; i++) {
    int maxn = 0;
    for (register int j = i + 1; j <= n; j++)
      maxn = max(maxn, pos[j] - pos[j - 1]), mx[i][j] = maxn;
  }
  random_shuffle(ints + 1, ints + 1 + m);
  long long maxn = 0;
  for (register int i = 1; i <= m; i++) {
    if (Tanxin(i, maxn)) continue;
    long long ans = ints[i].v * (long long)(pos[ints[i].t] - pos[ints[i].s]);
    long long l = max((long long)mx[ints[i].s][ints[i].t] * ints[i].v,
                      maxn + 1),
              r = ans;
    while (l <= r) {
      long long mid = (l + r) >> 1;
      if (Tanxin(i, mid))
        ans = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    maxn = max(maxn, ans);
  }
  cout << maxn << endl;
}
}  // namespace ywy
int main() {
  ywy::ywymain();
  return (0);
}