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;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 400;
const int MAXTRUCK = 250000;
int n, ntruck;
int x[MAXN];
int s[MAXTRUCK], t[MAXTRUCK], c[MAXTRUCK], r[MAXTRUCK];
int dp[MAXN][MAXN + 1];
int act[MAXN][MAXN + 1];
vector<int> bys[MAXN];
int calc(int i, int j, int k, int jj) {
  return max(dp[jj][k - 1], x[j] - x[jj]);
}
long long solve() {
  for (int i = (0); i < (n); ++i) bys[i].clear();
  for (int i = (0); i < (ntruck); ++i) bys[s[i]].push_back(i);
  long long ret = 0;
  for (int i = (0); i < (n); ++i) {
    for (int k = (0); k <= (n); ++k) dp[i][k] = 0, act[i][k] = i;
    for (int j = (i + 1); j < (n); ++j) dp[j][0] = x[j] - x[i], act[j][0] = i;
    for (int j = (i + 1); j < (n); ++j)
      for (int k = (1); k <= (n); ++k) {
        act[j][k] = act[j - 1][k];
        while (act[j][k] < j &&
               calc(i, j, k, act[j][k] + 1) < calc(i, j, k, act[j][k]))
          ++act[j][k];
        dp[j][k] = calc(i, j, k, act[j][k]);
      }
    for (int j = (0); j < (((int)(bys[i]).size())); ++j) {
      int idx = bys[i][j];
      long long cur = (long long)dp[t[idx]][r[idx]] * c[idx];
      ret = max(ret, cur);
    }
  }
  return ret;
}
void run() {
  scanf("%d%d", &n, &ntruck);
  for (int i = (0); i < (n); ++i) scanf("%d", &x[i]);
  for (int i = (0); i < (ntruck); ++i)
    scanf("%d%d%d%d", &s[i], &t[i], &c[i], &r[i]), --s[i], --t[i];
  cout << solve() << endl;
}
int main() {
  run();
  return 0;
}