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 long P = 1e9 + 7;
const int N = 401;
const long long INF = 1e16;
const double pi = acos(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int a[N], dp[N][N][N - 2];
int n, m, s, f, c, r;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (1); i <= (n); ++i) scanf("%d", a + i);
  for (int s = (1); s <= (n); ++s)
    for (int f = (s); f <= (n); ++f) dp[s][f][0] = a[f] - a[s];
  for (int r = (1); r <= (n - 2); ++r)
    for (int s = (1); s <= (n - r - 1); ++s)
      for (int f = (s + r + 1); f <= (n); ++f) {
        int le = s + r, ri = f;
        int &res = dp[s][f][r];
        res = dp[s][f][r - 1];
        while (le <= ri) {
          int mid = le + ri >> 1;
          int x = dp[s][mid][r - 1], y = a[f] - a[mid];
          if (x < y) {
            res = min(res, y);
            ri = mid - 1;
          } else {
            res = min(res, x);
            le = mid + 1;
          }
        }
      }
  long long ans = 0;
  for (int i = (1); i <= (m); ++i) {
    scanf("%d%d%d%d", &s, &f, &c, &r);
    ans = max(ans, 1ll * dp[s][f][min(r, f - s - 1)] * c);
  }
  cout << ans << endl;
  return 0;
}