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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class MarvaloGauntsRing {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] data = br.readLine().split(" ");
        String[] input = br.readLine().split(" ");

        long[] dataI = new long[data.length];
        long[] inputI = new long[input.length];
        for(int i=0;i<data.length;i++){
            dataI[i] = Integer.parseInt(data[i]);
        }
        for(int i=0;i<input.length;i++){
            inputI[i] = Integer.parseInt(input[i]);
        }

        int n = (int)dataI[0];
        long[] suffixMax = new long[n];
        long maxTillNow = (long) dataI[3] * inputI[n-1];
        for(int i=n-1;i>=0;i--){
            if(inputI[i]*dataI[3] > maxTillNow){
                maxTillNow = (long)inputI[i]*dataI[3];
            }
            suffixMax[i] = maxTillNow;
        }

        maxTillNow = inputI[0] * dataI[1];
        long max = Long.MIN_VALUE;
        for(int i=0;i<n;i++){
            if(inputI[i]*dataI[1] > maxTillNow){
                maxTillNow = inputI[i]*dataI[1];
            }
            long preMax = maxTillNow;
            long suffMax = suffixMax[i];
            long tempMax = preMax + suffMax + (dataI[2]*inputI[i]);
            if(tempMax > max){
                max = tempMax;
            }
        }

        System.out.println(max);
    }
}
