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;
const int maxn = 405;
const int inf = 0x3f3f3f3f;
int n, m;
int a[maxn];
int f[maxn][maxn][maxn];
template <typename T>
inline void minimize(T &target, const T &value) {
  if (target > value) {
    target = value;
  }
}
void divide(int a[], int f[], int g[], int l, int r, int pl, int pr) {
  if (l > r) {
    return;
  }
  int m = (l + r) / 2;
  int ptr = l;
  for (int i = pl; i < min(m, pr + 1); ++i) {
    int tmp = max(f[i], a[m] - a[i]);
    if (tmp < g[m]) {
      g[m] = tmp;
      ptr = i;
    }
  }
  divide(a, f, g, l, m - 1, pl, ptr);
  divide(a, f, g, m + 1, r, ptr, pr);
}
void compute(int a[], int n, int f[maxn][maxn]) {
  f[0][0] = 0;
  for (int i = 1; i < n; ++i) {
    divide(a, f[i - 1], f[i], 0, n - 1, 0, n - 1);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  memset(f, inf, sizeof(f));
  compute(a, n, f[0]);
  for (int i = 0; i < n; ++i) {
    compute(a + i, n - i, f[i]);
  }
  long long ans = 0;
  while (m--) {
    int s, f, c, r;
    scanf("%d%d%d%d", &s, &f, &c, &r), s--, f--;
    r++;
    r = min(r, f - s);
    r = max(r, 1);
    ans = max(ans, (long long)c * ::f[s][r][f - s]);
  }
  printf("%lld\n", ans);
  return 0;
}