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 maxn = 404, maxm = 250005;
int n, m, a[maxn];
long long dp[maxn][maxn][2], ans;
struct Truck {
  int id, s, f, c, r;
  long long ans;
} truck[maxm];
bool cmp1(Truck a, Truck b) {
  return a.r < b.r || a.r == b.r && a.s < b.s ||
         a.r == b.r && a.s == b.s && a.f < b.f;
}
bool cmp2(Truck a, Truck b) { return a.id < b.id; }
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d%d", &truck[i].s, &truck[i].f, &truck[i].c, &truck[i].r);
    truck[i].id = i;
    truck[i].r++;
    truck[i].r = min(truck[i].r, truck[i].f - truck[i].s);
  }
  sort(truck, truck + m, cmp1);
  int cnt = 0;
  memset(dp, 0x3f, sizeof(dp));
  for (int k = 1; k < n; k++) {
    for (int i = 1; i + k <= n; i++) {
      int l = i + k - 1;
      for (int j = i + k; j <= n; j++) {
        if (k == 1)
          dp[i][j][k % 2] = a[j] - a[i];
        else {
          while (l + 1 < j &&
                 max(dp[i][l][(k - 1) % 2], (long long)a[j] - a[l]) >
                     max(dp[i][l + 1][(k - 1) % 2], (long long)a[j] - a[l + 1]))
            l++;
          dp[i][j][k % 2] = max(dp[i][l][(k - 1) % 2], (long long)a[j] - a[l]);
        }
        while (cnt < m && truck[cnt].s == i && truck[cnt].f == j &&
               truck[cnt].r == k) {
          ans = max(ans, dp[i][j][k % 2] * truck[cnt++].c);
        }
      }
    }
  }
  sort(truck, truck + m, cmp2);
  printf("%lld\n", ans);
  return 0;
}