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;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
const int LOG = 21;
const int NN = 405;
const int MM = (int)2e4 + 25;
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9 + 10;
const int MAX_M = (int)3e5;
int mul(int x, int y) { return (1LL * x * y) % MOD; }
int add(int x, int y) {
  long long res = x;
  res += y;
  return (res % MOD + MOD) % MOD;
}
int fast_pow(int x, int y) {
  if (y == 0) return 1;
  if (y & 1)
    return mul(x, fast_pow(x, y - 1));
  else {
    int aux = fast_pow(x, y >> 1);
    return mul(aux, aux);
  }
}
long long two(int x) { return 1LL << x; }
bool contain(long long mask, int bit) { return (mask & two(bit)) > 0; }
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int bit_count(int x) { return x == 0 ? 0 : (x & 1) + bit_count(x >> 1); }
int n;
int a[NN];
int dp[NN][NN][NN];
int main() {
  int m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int s = 1; s <= n; s++) {
    for (int f = s; f <= n; f++) dp[s][f][0] = a[f] - a[s];
  }
  for (int rem = 1; rem <= n; rem++) {
    for (int l = 1; l <= n; l++) {
      int idx = l;
      for (int r = l + 1; r <= n; r++) {
        while (dp[l][idx][rem - 1] <= a[r] - a[idx]) idx++;
        dp[l][r][rem] = INF;
        for (int z = -1; z <= 0; z++)
          if (l <= z + idx && z + idx <= r) {
            int mid = z + idx;
            dp[l][r][rem] =
                ((dp[l][r][rem]) < (((a[r] - a[mid]) > (dp[l][mid][rem - 1])
                                         ? (a[r] - a[mid])
                                         : (dp[l][mid][rem - 1])))
                     ? (dp[l][r][rem])
                     : (((a[r] - a[mid]) > (dp[l][mid][rem - 1])
                             ? (a[r] - a[mid])
                             : (dp[l][mid][rem - 1]))));
          }
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < m; i++) {
    int s, f, c, r;
    scanf("%d%d%d%d", &s, &f, &c, &r);
    ans = ((ans) > (1LL * c * dp[s][f][r]) ? (ans) : (1LL * c * dp[s][f][r]));
  }
  printf("%lld\n", ans);
  return 0;
}