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;
struct node {
  int j, k;
  long long c;
};
constexpr int M = 3 * 1e5, N = 435;
constexpr long long INF = 1000000000000000000ll;
int n, m, big[N][N], cut[N][N];
long long dp[N][N], d[N], res;
vector<node> v[N];
inline node make_node(int a, int b, long long _c) {
  node ret;
  ret.j = a;
  ret.k = b;
  ret.c = _c;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> d[i], big[i - 1][i] = d[i] - d[i - 1];
  for (int i = 0; i < m; i++) {
    int s, f, c, r;
    cin >> s >> f >> c >> r;
    v[s].push_back(make_node(f, r, c));
  }
  for (int i = 0; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      big[i][j] = big[j][i] = max(d[j] - d[j - 1], (long long)big[i][j - 1]);
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      for (int k = 0; k <= n; k++) {
        dp[j][k] = INF;
        cut[j][k] = -1;
      }
      dp[j][0] = d[j] - d[i];
      cut[j][0] = 0;
    }
    for (int j = 0; j <= i; j++) {
      for (int k = 0; k <= n; k++) {
        dp[j][k] = 0;
      }
    }
    for (int i = 0; i <= n; i++) {
      cut[n + 1][i] = n + 1;
    }
    for (int len = 1; len <= n - i; len++) {
      for (int j = n; j >= i; j--) {
        dp[j][len] = dp[j][len - 1];
        cut[j][len] = cut[j][len - 1];
        for (int k = max(cut[j][len - 1], i); k <= min(cut[j + 1][len], j);
             k++) {
          if (dp[j][len] > max(dp[k][len - 1], d[j] - d[k])) {
            cut[j][len] = k;
            dp[j][len] = max(dp[k][len - 1], d[j] - d[k]);
          }
        }
      }
    }
    for (node nn : v[i]) {
      res = max(res, nn.c * dp[nn.j][min(nn.j - i - 1, nn.k)]);
    }
  }
  cout << res << '\n';
}