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 int N = 3e5 + 10;
const int INF = 0x3f3f3f3f;
int dp[405][405], s[N], f[N], c[N], r[N], a[405], n, m;
long long ans;
vector<int> V[405];
int main() {
  scanf("%d %d", &n, &m);
  long long ans = 0;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= m; i++)
    scanf("%d %d %d %d", &s[i], &f[i], &c[i], &r[i]), V[s[i]].push_back(i);
  for (int i = 1; i <= n; i++) {
    memset(dp, 0x3f, sizeof dp);
    dp[i][0] = 0;
    for (int j = i + 1; j <= n; j++) {
      for (int k = 1; k <= n; k++) {
        if (k == j - i)
          dp[j][k] = max(dp[j - 1][k - 1], a[j] - a[j - 1]);
        else if (k > j - i)
          dp[j][k] = dp[j][k - 1];
        else {
          int l = i + k - 1, r = j - 1, ans = -1;
          while (l <= r) {
            int mid = l + r >> 1;
            if (dp[mid][k - 1] >= a[j] - a[mid]) {
              ans = mid;
              r = mid - 1;
            } else
              l = mid + 1;
          }
          if (ans == -1)
            dp[j][k] = a[j] - a[j - 1];
          else
            dp[j][k] = dp[ans][k - 1];
          if (ans > i) dp[j][k] = min(dp[j][k], a[j] - a[ans - 1]);
        }
      }
    }
    for (auto it : V[i]) {
      long long sum;
      if (r[it] == 0)
        sum = (a[f[it]] - a[s[it]]) * 1ll * c[it];
      else if (r[it] >= f[it] - s[it])
        sum = dp[f[it]][r[it]] * 1ll * c[it];
      else {
        sum = INF;
        for (int j = s[it] + r[it]; j <= f[it]; j++) {
          sum = min(sum, 0ll + max(dp[j][r[it]], a[f[it]] - a[j]));
        }
        sum = sum * 1ll * c[it];
      }
      ans = max(ans, sum);
    }
  }
  cout << ans << endl;
  return 0;
}