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;
inline long long read() {
  long long x = 0, neg = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return x * neg;
}
inline void print(long long x) {
  if (x < 0) {
    putchar('-');
    print(abs(x));
    return;
  }
  if (x <= 9)
    putchar(x + '0');
  else {
    print(x / 10);
    putchar(x % 10 + '0');
  }
}
inline long long qpow(long long x, long long e, long long _MOD) {
  long long ans = 1;
  while (e) {
    if (e & 1) ans = ans * x % _MOD;
    x = x * x % _MOD;
    e >>= 1;
  }
  return ans;
}
long long dp[405][405][2], a[250005], b[250005], c[250005], d[250005],
    n = read(), m = read(), t[405];
vector<long long> q[250005];
long long ans = 0;
signed main() {
  for (long long i = 1; i <= n; i++) t[i] = read();
  for (long long i = 1; i <= m; i++) {
    a[i] = read();
    b[i] = read();
    c[i] = read();
    d[i] = read();
    d[i]++;
    d[i] = min(d[i], b[i] - a[i]);
    q[d[i]].push_back(i);
  }
  memset(dp, 0x3f, sizeof(dp));
  for (long long i = 1; i <= n; i++)
    for (long long j = i; j <= n; j++) dp[i][j][1] = t[j] - t[i];
  for (long long k = 1; k <= n - 1; k++) {
    for (__typeof(q[k].begin()) it = q[k].begin(); it != q[k].end(); it++) {
      ans = max(ans, dp[a[*it]][b[*it]][k & 1] * c[*it]);
    }
    for (long long i = 1; i <= n; i++) {
      dp[i][i][(k + 1) & 1] = 0;
      long long l = i;
      for (long long j = i + 1; j <= n; j++) {
        while (dp[i][l][k & 1] < t[j] - t[l]) l++;
        dp[i][j][(k + 1) & 1] = min(dp[i][l][k & 1], t[j] - t[l - 1]);
      }
    }
  }
  cout << ans << endl;
  return 0;
}