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 JAVA):

Solution:
// upsolve with rainboy
import java.io.*;
import java.util.*;

public class CF1101F {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int m = Integer.parseInt(st.nextToken());
		st = new StringTokenizer(br.readLine());
		int[] aa = new int[n];
		for (int i = 0; i < n; i++)
			aa[i] = Integer.parseInt(st.nextToken());
		int[] ss = new int[m];
		int[] ff = new int[m];
		int[] cc = new int[m];
		int[] rr = new int[m];
		Integer[] ii = new Integer[m];
		for (int h = 0; h < m; h++) {
			st = new StringTokenizer(br.readLine());
			ss[h] = Integer.parseInt(st.nextToken()) - 1;
			ff[h] = Integer.parseInt(st.nextToken()) - 1;
			cc[h] = Integer.parseInt(st.nextToken());
			rr[h] = Integer.parseInt(st.nextToken());
			ii[h] = h;
		}
		Arrays.sort(ii, (i, j) -> { return rr[i] - rr[j]; });
		int[][] dp = new int[n][n];
		int[][] dq = new int[n][n];
		long v = 0;
		for (int s = 0; s < n; s++)
			for (int f = s + 1; f < n; f++)
				dp[s][f] = aa[f] - aa[s];
		for (int r = 0, h_ = 0, h; r <= n; r++) {
			while (h_ < m && rr[h = ii[h_]] == r) {
				v = Math.max(v, (long) dp[ss[h]][ff[h]] * cc[h]);
				h_++;
			}
			for (int s = 0; s < n; s++) {
				int i = s;
				for (int f = s + 1; f < n; f++) {
					while (i < f && dp[s][i] < aa[f] - aa[i])
						i++;
					dq[s][f] = Math.min(dp[s][i], aa[f] - aa[i - 1]);
				}
			}
			int[][] tmp = dp; dp = dq; dq = tmp;
		}
		System.out.println(v);
	}
}