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.util.*;
public class Main {
	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		int n=in.nextInt();
		int m=in.nextInt();
		int[] a=new int[n];
		int[] s=new int[m];
		int[] f=new int[m];
		int[] c=new int[m];
		int[] r=new int[m];
		for (int i=0;i<n;++i) a[i]=in.nextInt();
		for (int i=0;i<m;++i) {
			s[i]=in.nextInt();
			f[i]=in.nextInt();
			c[i]=in.nextInt();
			r[i]=in.nextInt();
			--s[i];
			--f[i];
		}
		// dp[i][j][k]=max(dp[i][l][k-1]+a[j]-a[l]);
		int[][][] dp=new int[n][n][n];
		int inf=0x3f3f3f3f;
		for (int i=0;i<n;++i) {
			for (int j=i;j<n;++j) dp[i][j][0]=a[j]-a[i];
			for (int k=1;k<n;++k) {
				int l=i;
				for (int j=i+k+1;j<n;++j) {
					while (l+1<j&&Math.max(dp[i][l][k-1],a[j]-a[l])>Math.max(dp[i][l+1][k-1],a[j]-a[l+1]))
						++l;
					dp[i][j][k]=Math.max(dp[i][l][k-1],a[j]-a[l]);
				}
			}
		}
		long res=0;
		for (int i=0;i<m;++i) res=Math.max(res,(long)c[i]*dp[s[i]][f[i]][Math.min(r[i],f[i]-s[i]-1)]);
		System.out.println(res);
	}
}