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 N = 405;
const int inf = 1e9 + 7;
int dp[N][N][N], a[N];
int DP(int l, int r, int c) {
  if (l == r) return 0;
  if (l + 1 == r) return a[r] - a[l];
  if (dp[l][r][c]) return dp[l][r][c];
  if (c == 0) return dp[l][r][c] = a[r] - a[l];
  int opt = (a[r] - a[l]) / (c + 1);
  int i = lower_bound(a + l, a + r + 1, a[l] + opt) - a;
  dp[l][r][c] = inf;
  if (i != l)
    dp[l][r][c] = min(dp[l][r][c], max(DP(i - 1, r, c - 1), a[i - 1] - a[l]));
  dp[l][r][c] = min(dp[l][r][c], max(DP(i, r, c - 1), a[i] - a[l]));
  if (i != r)
    dp[l][r][c] = min(dp[l][r][c], max(DP(i + 1, r, c - 1), a[i + 1] - a[l]));
  return dp[l][r][c];
}
int main() {
  int n, m, i;
  scanf("%i %i", &n, &m);
  for (i = 1; i <= n; i++) scanf("%i", &a[i]);
  long long ans = 0;
  int l, r, mul, c;
  while (m--) {
    scanf("%i %i %i %i", &l, &r, &mul, &c);
    c = min(c, r - l);
    ans = max(ans, (long long)DP(l, r, c) * mul);
  }
  printf("%lld\n", ans);
  return 0;
}