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>
namespace OI {
template <class T>
T gcd(T x, T y) {
  return !y ? x : gcd(y, x % y);
}
template <class T>
T Fabs(T x) {
  return x < 0 ? -x : x;
}
template <class T>
void rd(T &x) {
  x = 0;
  int f = 1;
  char c;
  while (!isdigit(c = getchar()))
    if (c == '-') f = -1;
  do {
    x = (x << 3) + (x << 1) + (c ^ 48);
  } while (isdigit(c = getchar()));
  x *= f;
}
template <class T>
void pt(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) pt(x / 10);
  putchar(x % 10 ^ 48);
}
}  // namespace OI
using namespace OI;
using namespace std;
const int N = 401;
int n, m, a[N], dp[N][N][N];
long long ans;
int main() {
  rd(n), rd(m);
  for (int i = 1; i <= n; i++) rd(a[i]);
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) dp[i][j][0] = a[j] - a[i];
  for (int i = 1; i <= n; i++)
    for (int k = 1; k <= n; k++)
      for (int j = i + 1, t = i; j <= n; j++) {
        dp[i][j][k] = dp[i][j][k - 1];
        while (dp[i][t][k - 1] <= a[j] - a[t] && t < j) t++;
        dp[i][j][k] = min(dp[i][j][k], min(dp[i][t][k - 1], a[j] - a[t - 1]));
      }
  while (m--) {
    int s, t, v, r;
    rd(s), rd(t), rd(v), rd(r);
    r = min(n, r);
    ans = max(ans, 1ll * dp[s][t][r] * v);
  }
  pt(ans);
  return 0;
}