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;
namespace IO {
void setIn(string s) { freopen(s.c_str(), "r", stdin); }
void setOut(string s) { freopen(s.c_str(), "w", stdout); }
void setIO(string s = "") {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  if (s.size()) {
    setIn(s + ".inp");
    setOut(s + ".out");
  } else {
  }
}
}  // namespace IO
using namespace IO;
namespace Function {
template <typename T1, typename T2>
void amax(T1 &a, T2 b) {
  assert(!(typeid(a).name() == typeid(int).name() &&
           typeid(b).name() == typeid(long long).name()));
  if (a < b) a = b;
}
template <typename T1, typename T2>
void amin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T>
void compress(T &a) {
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
}
template <typename T1, typename T2, typename T3>
int position(T1 Begin, T2 End, T3 val, bool type = 0) {
  if (type == 0) {
    return lower_bound(Begin, End, val) - Begin;
  }
  return upper_bound(Begin, End, val) - Begin;
}
template <typename T>
long long sqr(T x) {
  return 1LL * x * x;
}
template <typename T1, typename T2>
long long GCD(T1 a, T2 b) {
  return b == 0 ? a : GCD(b, a % b);
}
template <typename T1, typename T2>
long long LCM(T1 a, T2 b) {
  return 1LL * a / GCD(a, b) * b;
}
}  // namespace Function
using namespace Function;
namespace Output {
void print(int x) { cout << x << "\n"; }
void print(unsigned int x) { cout << x << "\n"; }
void print(long unsigned int x) { cout << x << "\n"; }
void print(long long x) { cout << x << "\n"; }
void print(unsigned long long x) { cout << x << "\n"; }
void print(float x) { cout << x << "\n"; }
void print(double x) { cout << x << "\n"; }
void print(long double x) { cout << x << "\n"; }
void print(char x) { cout << x << "\n"; }
void print(char *x) { cout << x << "\n"; }
void print(unsigned char x) { cout << x << "\n"; }
void print(const char *x) { cout << x << "\n"; }
void print(string x) { cout << x << "\n"; }
void print(bool x) { cout << x << "\n"; }
template <class T, class... Ts>
void print(T t, Ts... ts) {
  cout << t << " ";
  print(ts...);
}
template <typename T1, typename T2>
void print(pair<T1, T2> a) {
  print(a.first, a.second);
}
template <typename T>
void print(T a) {
  for (auto it : a) {
    print(it);
  }
}
template <class T, class... Ts>
void prine(T t, Ts... ts) {
  print(t, ts...);
  exit(0);
}
}  // namespace Output
using namespace Output;
const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};
const int INF = 1e9;
const long long INFL = 1e18;
const int MOD = 1e9 + 7;
const int N = 4e2 + 10;
const int M = 25e4 + 10;
int dp[N][N];
int a[N];
int s[M], f[M], c[M], r[M];
vector<int> G[M];
int main() {
  setIO();
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < m; i++) {
    cin >> s[i] >> f[i] >> c[i] >> r[i];
    G[s[i]].push_back(i);
  }
  long long res = 0;
  for (int l = 1; l <= n; l++) {
    for (int r = l; r <= n; r++) {
      dp[0][r] = a[r] - a[l];
    }
    for (int i = 1; i <= n; i++) {
      int opt = l;
      for (int r = l; r <= n; r++) {
        while (opt < r && max(dp[i - 1][opt], a[r] - a[opt]) >=
                              max(dp[i - 1][opt + 1], a[r] - a[opt + 1])) {
          opt++;
        }
        dp[i][r] = max(dp[i - 1][opt], a[r] - a[opt]);
      }
    }
    for (int x : G[l]) {
      amax(res, 1LL * dp[r[x]][f[x]] * c[x]);
    }
  }
  print(res);
  return 0;
}