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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class F {
	
	static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }
	static class Node
	{
		int s;
		int f;
		int c;
		int r;
		Node(int s,int f,int c,int r)
		{
			this.s = s;
			this.f = f;
			this.c = c;
			this.r = r;
		}
	}
	static int a[];
	static boolean check(long ans,Node p)
	{	
		long f= ans;
        int c=0;
        for(int j=p.s+1;j<=p.f;j++){
            if(f-(long)(a[j]-a[j-1])*p.c<0){
                return false;
            }
            f-=(long)(a[j]-a[j-1])*p.c;
            if(j<p.f && f-(long)(a[j+1]-a[j])*p.c<0){
                f= ans;
                c++;
                if(c>p.r){
                    return false;
                }
            }
        }
        return true;
	}
	static long search(Node p){
        long l=1,r=1000000000;
        r*=r;
        long ans= 0;
        while(l<=r){
            long m=(l+r)>>1;
            if(check(m,p)){
                ans=m;
                r=m-1;
            }else l=m+1;
        }
        return ans;
    }
	public static void main(String[] args) 
	{
		OutputStream outputStream = System.out;
        FastReader sc = new FastReader();
        PrintWriter out = new PrintWriter(outputStream);
        int n = sc.nextInt();
        int m = sc.nextInt();
        a = new int[n+1];
        for(int i = 1; i <= n; i++)
        	a[i] = sc.nextInt();
        
        ArrayList<Node> ar = new ArrayList<>();
        for(int i = 0; i < m; i++)
        	ar.add(new Node(sc.nextInt(),sc.nextInt(),sc.nextInt(),sc.nextInt()));
        
        Collections.shuffle(ar);
        long ans = 0;
        for(Node p: ar)
        {
        	if(!check(ans,p))
        	{
        		ans = search(p);
        	}
        }
        out.println(ans);
        out.close();
	}
	
}