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 = 400 + 5;
const int mo = 998244353;
long long readint() {
  long long x = 0, f = 1;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    (s = getchar());
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s ^ 48);
    (s = getchar());
  }
  return x * f;
}
long long ind[maxn];
struct zz {
  long long s, t, c, r;
  long long divide;
} a[250005];
long long dp[maxn][maxn][2];
vector<long long> vec[maxn];
signed main() {
  long long n = readint(), m = readint();
  for (long long i = 1; i <= n; i++) {
    ind[i] = readint();
  }
  for (long long i = 1; i <= m; i++) {
    a[i].s = readint(), a[i].t = readint();
    a[i].c = readint(), a[i].r = readint() + 1;
    a[i].divide = min(a[i].r, a[i].t - a[i].s);
    vec[a[i].divide].push_back(i);
  }
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= n; j++)
      for (long long k = 0; k <= 1; k++) dp[i][j][k] = 0x3f3f3f3f;
  for (long long i = 1; i <= n; i++)
    for (long long j = i + 1; j <= n; j++) {
      dp[i][j][1] = ind[j] - ind[i];
    }
  long long ans = 0;
  for (long long i = 0; i < vec[1].size(); i++) {
    ans =
        max(ans, 1ll * dp[a[vec[1][i]].s][a[vec[1][i]].t][1] * a[vec[1][i]].c);
  }
  for (long long k = 2; k <= n; k++) {
    for (long long l = 1; l <= n; l++) {
      dp[l][l][k & 1] = 0;
      long long i = l + k - 1;
      for (long long r = l + k; r <= n; r++) {
        while (dp[l][i][(k - 1) & 1] < ind[r] - ind[i]) i++;
        dp[l][r][k & 1] = min(dp[l][i][(k - 1) & 1], ind[r] - ind[i - 1]);
      }
    }
    for (long long j = 0; j < vec[k].size(); j++) {
      ans = max(ans, 1ll * dp[a[vec[k][j]].s][a[vec[k][j]].t][k & 1] *
                         a[vec[k][j]].c);
    }
  }
  cout << ans << endl;
  return 0;
}