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 cf1101f {

	public static void main(String[] args)throws IOException {
		BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		StringTokenizer st=new StringTokenizer(bf.readLine());
		int n=Integer.parseInt(st.nextToken());
		int m=Integer.parseInt(st.nextToken());
		st=new StringTokenizer(bf.readLine());
		int[] a=new int[n];
		for(int i=0;i<n;i++) {
			a[i]=Integer.parseInt(st.nextToken());
		}
		int[] s=new int[m];
		int[] f=new int[m];
		int[] c=new int[m];
		pair[] r=new pair[m];
		for(int i=0;i<m;i++) {
			st=new StringTokenizer(bf.readLine());
			s[i]=Integer.parseInt(st.nextToken())-1;
			f[i]=Integer.parseInt(st.nextToken())-1;
			c[i]=Integer.parseInt(st.nextToken());
			r[i]=new pair(Integer.parseInt(st.nextToken()),i);
		}
		Arrays.sort(r);
		int[][] dp=new int[n][n];
		int[][] dp2=new int[n][n];
		long v=0;
		for(int i=0;i<n;i++) {
			for(int j=i+1;j<n;j++) {
				dp[i][j]=a[j]-a[i];
			}
		}
		int cur=0;
		for(int i=0;i<=n;i++) {
			while(cur<m&&r[cur].a==i) {
				int x=r[cur].b;
				v=Math.max(v, (long)dp[s[x]][f[x]]*c[x]);
				cur++;
			}
			for(int j=0;j<n;j++) {
				int l=j;
				for(int k=j+1;k<n;k++) {
					while(l<k&&dp[j][l]<a[k]-a[l]) {
						l++;
					}
					dp2[j][k]=Math.min(dp[j][l], a[k]-a[l-1]);
				}
			}
			int[][] swap=dp;
			dp=dp2;
			dp2=swap;
		}
		out.println(v);
		out.close();
	}
	static class pair implements Comparable<pair>{
		int a;int b;
		public pair(int x, int y) {
			a=x;b=y;
		}
		public int compareTo(pair p) {
			return a-p.a;
		}
	}
}