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;
template <class U, class T>
void Max(U &first, T second) {
  if (first < second) first = second;
}
template <class U, class T>
void Min(U &first, T second) {
  if (first > second) first = second;
}
template <class T>
void add(int &a, T b) {
  a = (a + b) % 1000000007;
}
inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % 1000000007;
    a = 1LL * a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
int pow(int a, int b, int c) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % c;
    a = 1LL * a * a % c;
    b >>= 1;
  }
  return ans;
}
struct node {
  int s, t, c, r;
} e[250010];
int a[401];
int check(node &e, long long w) {
  long long s = w, r = e.r, v;
  for (int i = e.s; i < e.t; i++) {
    v = 1LL * e.c * a[i];
    if (s < v) s = w, r--;
    s -= v;
    if (s < 0 || r < 0) return 0;
  }
  return 1;
}
int main() {
  int ca = 0, T, k, i, j, m = 0, K, n;
  double start = clock();
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n - 1; i++) a[i] = a[i + 1] - a[i];
  a[n - 1] = 0;
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d%d", &e[i].s, &e[i].t, &e[i].c, &e[i].r);
    e[i].s--, e[i].t--;
  }
  random_shuffle(e, e + m);
  long long ans = 0;
  for (int i = 0; i < m; i++)
    if (!check(e[i], ans)) {
      long long l = ans + 1, r = 1e18;
      while (l <= r) {
        long long w = (l + r) >> 1;
        if (check(e[i], w))
          r = w - 1;
        else
          l = w + 1;
      }
      ans = r + 1;
    }
  printf("%lld\n", ans);
  cerr << (1. * clock() - start) / CLOCKS_PER_SEC << "\n";
}