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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
int n, m;
long long dp[2][404][404], a[404];
vector<tuple<int, int, int> > q[404];
inline long long calc(int k, int i, int p, int j) {
  return max(dp[k & 1][i][p], a[j] - a[p]);
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i <= m; ++i) {
    int s = read(), t = read(), c = read(), r = read();
    q[r + 1].emplace_back(s, t, c);
  }
  memset(dp[0], 0x3f, sizeof(dp[0]));
  for (int i = 1; i <= n; ++i) {
    dp[0][i][i] = 0;
  }
  long long ans = 0;
  for (int k = 1; k <= n + 1; ++k) {
    for (int i = 1; i <= n; ++i) {
      int p = i;
      dp[k & 1][i][i] = 0;
      for (int j = i + 1; j <= n; ++j) {
        while (p < j - 1 && calc(k - 1, i, p + 1, j) <= calc(k - 1, i, p, j))
          ++p;
        dp[k & 1][i][j] = calc(k - 1, i, p, j);
      }
    }
    for (auto [s, t, c] : q[k]) {
      ans = max(ans, dp[k & 1][s][t] * c);
    }
  }
  printf("%lld\n", ans);
  return 0;
}