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.*;
import java.util.*;

public class B{

	static long INF=(long)1e18;
	public static void main(String[] args) throws IOException {
		Scanner sc=new Scanner();
		PrintWriter out=new PrintWriter(System.out);
		int n=sc.nextInt(),m=sc.nextInt();
		int []a=new int [n];
		for(int i=0;i<n;i++)
			a[i]=sc.nextInt();
		long ans=1;
		int []indices=new int [m],S=new int [m],F=new int [m],C=new int [m],R=new int [m];
		for(int i=0;i<m;i++) {
			indices[i]=i;
			S[i]=sc.nextInt()-1;
			F[i]=sc.nextInt()-1;
			C[i]=sc.nextInt();
			R[i]=sc.nextInt();
		}
		Random rand =new Random();
		for(int i=0;i<m;i++) {
			int r=rand.nextInt(m);
			
//			System.out.println(r);
			int tmp=indices[r];
			indices[r]=indices[i];
			indices[i]=tmp;
		}
				for(int j=0;j<m;j++) {
//			int s=sc.nextInt()-1,f=sc.nextInt()-1,c=sc.nextInt(),r=sc.nextInt();
			int idx=indices[j];
			int s=S[idx],f=F[idx],c=C[idx],r=R[idx];
			long start=ans;
				boolean ok=true;
				long fuel=start;
				int refuel=r;
				for(int i=s+1;i<=f && ok;i++) {
					long needed=(a[i]-a[i-1])*1l*c;
					fuel-=needed;
					if(fuel<0) {
						fuel=start-needed;
						if(refuel--==0 || fuel<0)
							ok=false;
					}
				}
				if(ok) {
					continue;
				}
				
				
			
			long lo=ans,hi=INF,curr=INF;
			while(lo<=hi) {
				 start=lo+hi>>1;
				 ok=true;
				 fuel=start;
				 refuel=r;
				for(int i=s+1;i<=f && ok;i++) {
					long needed=(a[i]-a[i-1])*1l*c;
					fuel-=needed;
					if(fuel<0) {
						fuel=start-needed;
						if(refuel--==0 || fuel<0)
							ok=false;
					}
				}
				if(ok) {
					curr=start;
					hi=start-1;
				}
				else
					lo=start+1;
				
			}
			ans=Math.max(ans, curr);
		
		}
		out.println(ans);
		out.close();

	}
	static class pair{
		int l,r,idx;
		pair(int a,int b,int c){
			l=a;
			r=b;
			idx=c;
		}
	}
	static class Scanner
	{
		BufferedReader br;
		StringTokenizer st;
		Scanner(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		Scanner(String fileName) throws FileNotFoundException{
			br=new BufferedReader(new FileReader(fileName));
		}
		String next() throws IOException {
			while(st==null || !st.hasMoreTokens())
				st=new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		String nextLine() throws IOException {
			return br.readLine();
		}
		int nextInt() throws IOException{
			return Integer.parseInt(next());
		}
		long nextLong()  throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}
		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}
	}
}