Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly x drops of the potion he made. 

Value of x is calculated as maximum of p·ai + q·aj + r·ak for given p, q, r and array a1, a2, ... an such that 1 ≤ i ≤ j ≤ k ≤ n. Help Snape find the value of x. Do note that the value of x may be negative.

Input

First line of input contains 4 integers n, p, q, r ( - 109 ≤ p, q, r ≤ 109, 1 ≤ n ≤ 105).

Next line of input contains n space separated integers a1, a2, ... an ( - 109 ≤ ai ≤ 109).

Output

Output a single integer the maximum value of p·ai + q·aj + r·ak that can be obtained provided 1 ≤ i ≤ j ≤ k ≤ n.

Examples

Input

5 1 2 3
1 2 3 4 5


Output

30


Input

5 1 2 -3
-1 -2 -3 -4 -5


Output

12

Note

In the first sample case, we can take i = j = k = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.

In second sample case, selecting i = j = 1 and k = 5 gives the answer 12.
import java.util.*;

public class B {

   public static void main (String [] args) {
   
      Scanner sc = new Scanner(System.in);
      int N = sc.nextInt();
      long [] c = new long [3];
      for (int k=0; k<3; k++)
         c[k]=sc.nextLong();
      long [] a = new long[N];
      for (int k=0; k<N; k++)
         a[k]=sc.nextLong();
      long [][] dp = new long[3][N];
      dp[0][0]=c[0]*a[0];
      for (int k=1; k<N; k++)
         dp[0][k]=Math.max(dp[0][k-1], c[0]*a[k]);
      //System.out.println(Arrays.toString(dp[0]));
      for (int k=1; k<3; k++) {
         dp[k][0]=c[k]*a[0]+dp[k-1][0];
         for (int j=1; j<N; j++) 
            dp[k][j]=Math.max(dp[k][j-1], dp[k-1][j]+c[k]*a[j]);
         //System.out.println(Arrays.toString(dp[k]));       
      }   
      System.out.println(dp[2][N-1]);
   }
}