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>
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
#pragma GCC target(sse, sse2, sse3, ssse3, sse4, popcnt, tune = native)
#pragma GCC optimize(                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           \
    "-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2", \
    3)
using namespace std;
const long long MAXN = 447;
long long n, m;
vector<long long> city;
vector<pair<pair<long long, long long>, pair<long long, long long> > > car;
void read() {
  cin >> n >> m;
  city.resize(n);
  for (long long i = 0; i < n; i++) cin >> city[i];
  car.resize(m);
  for (long long i = 0; i < m; i++) {
    cin >> car[i].first.second >> car[i].second.first >> car[i].second.second >>
        car[i].first.first;
  }
}
long long dp[2][MAXN][MAXN], op[2][MAXN][MAXN];
long long get_hull(long long &l, long long &r, long long z) {
  long long res = 1e17 + 47;
  for (long long opt = op[z][l][r]; opt < r + 1; opt++) {
    res = min(res, max(city[r] - city[opt], dp[z][l][opt]));
    if (city[r] - city[opt] < dp[z][l][opt]) {
      op[z ^ 1][l][r] = opt;
      return res;
    }
  }
  op[z ^ 1][l][r] = r;
  return res;
}
bool good(pair<pair<long long, long long>, pair<long long, long long> > &a,
          long long z, long long &l, long long &r) {
  if (a.first.first == z && a.first.second - 1 == l && a.second.first - 1 == r)
    return true;
  return false;
}
void solve() {
  sort(car.begin(), car.end());
  reverse(car.begin(), car.end());
  for (long long l = 0; l < n; l++)
    for (long long r = l; r < n; r++) dp[0][l][r] = city[r] - city[l];
  long long ans = 0;
  for (long long l = 0; l < n; l++) {
    for (long long r = l; r < n; r++) {
      while ((long long)(car.size()) > 0 && good(car.back(), 0, l, r))
        ans = max(ans, car.back().second.second * dp[0][l][r]), car.pop_back();
    }
  }
  for (long long z = 1; z < n + 1; z++) {
    long long kek = z % 2;
    for (long long l = 0; l < n; l++) {
      memset(dp[kek][l], 0, sizeof(dp[kek][l]));
      memset(op[kek][l], 0, sizeof(op[kek][l]));
      for (long long r = l; r < n; r++) {
        dp[kek][l][r] = get_hull(l, r, kek ^ 1);
        while ((long long)(car.size()) > 0 && good(car.back(), z, l, r))
          ans = max(ans, car.back().second.second * dp[kek][l][r]),
          car.pop_back();
      }
    }
  }
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  read();
  solve();
}