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:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.*;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.Comparator;
import java.util.ArrayList;
import java.util.concurrent.ThreadLocalRandom;
import java.util.Collections;

public class Main {
    public static void main(String[] args) throws IOException  {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(in, out);
        out.close();
    }
    static class Truck{
        private int st, fin, consumption , cntRefueling;
        Truck (int st, int fin, int consumption, int cntRefueling){
            this.st = st;
            this.fin = fin;
            this.consumption = consumption;
            this.cntRefueling = cntRefueling;
        }
        public boolean check(long tankVolume, long[] arr){
            long cntRefCopy = cntRefueling, volumeNow = tankVolume;
            boolean ans = true;
            for(int i = st - 1; i < fin - 1; ++i){
                long tmp = consumption * (arr[i + 1] - arr[i]);
                if(tmp > tankVolume) {ans = false; break;}
                if(tmp > volumeNow){
                    if(cntRefCopy > 0){
                        cntRefCopy--;
                        volumeNow = tankVolume - tmp;
                    }
                    else {ans = false; break;}
                }
                else volumeNow -= tmp;
            }
            return ans;
        }
    }
    static class TaskA {
        long[] arr;
        Truck[] def;
        public long binSearchM(long left, long right, int indx){
            long midd = 0;
            while(left + 1 < right) {
                midd = (left + right) / 2;
                if(def[indx].check(midd, arr))
                    right = midd;
                else
                    left = midd ;
            }
            //while(!def[indx].check(midd, arr)) midd++;
            return right;
        }
        static void shuffleArray(Truck[] ar)
        {
            Random rnd = ThreadLocalRandom.current();
            for (int i = ar.length - 1; i > 0; i--) {
                int index = rnd.nextInt(i + 1);
                Truck a = ar[index];
                ar[index] = ar[i];
                ar[i] = a;
            }
        }
        public void solve(InputReader in, PrintWriter out) throws IOException  {
            int n = in.nextInt(), m = in.nextInt();
            arr  = new long[n];
            def = new Truck[m];
            for(int i = 0; i < n; ++i)
                arr[i] = in.nextLong();
            for(int i = 0; i < m; ++i)
                def[i] = new Truck(in.nextInt(), in.nextInt(), in.nextInt(), in.nextInt());
            shuffleArray(def);
            long minV = 0;
            for(int i = 0; i < m; ++i){
                if(!def[i].check(minV, arr))
                    minV = binSearchM(minV, (long)1e18, i);

            }
            out.println(minV);
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

    }
}