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 n, m;
int K;
int x, y, xx, yy;
deque<int> q[505];
int dp[505][505], dp2[505];
long long ans;
int a[505];
vector<pair<pair<int, int>, int> > qq[505];
void solve(int l) {
  for (int i = 0; i <= K; ++i) {
    dp[l][i] = 0;
    q[i].clear();
    q[i].push_back(l);
    dp2[i] = -1e9;
  }
  int cur, x;
  for (int i = l + 1; i <= n; ++i) {
    for (int j = 1; j <= K; ++j) {
      while (!q[j - 1].empty()) {
        x = q[j - 1].front();
        if (dp[x][j - 1] <= a[i] - a[x]) {
          q[j - 1].pop_front();
          dp2[j - 1] = a[x];
        } else
          break;
      }
      cur = a[i] - dp2[j - 1];
      if (!q[j - 1].empty()) {
        x = q[j - 1].front();
        if (dp[x][j - 1] < cur) cur = dp[x][j - 1];
      }
      dp[i][j] = cur;
      q[j].push_back(i);
    }
  }
  int y, xx;
  long long cur2;
  for (int i = 0; i < qq[l].size(); ++i) {
    x = qq[l][i].first.first;
    y = qq[l][i].first.second;
    xx = qq[l][i].second;
    xx++;
    cur2 = dp[x][xx];
    cur2 *= y;
    if (cur2 > ans) ans = cur2;
  }
}
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  K = n + 2;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= m; ++i) {
    cin >> x >> y >> xx >> yy;
    qq[x].push_back(make_pair(make_pair(y, xx), yy));
  }
  for (int i = 1; i <= n; ++i) solve(i);
  cout << ans << '\n';
}