One day Vasya got hold of information on the Martian dollar course in bourles for the next n days. The buying prices and the selling prices for one dollar on day i are the same and are equal to ai. Vasya has b bourles. He can buy a certain number of dollars and then sell it no more than once in n days. According to Martian laws, one can buy only an integer number of dollars. Which maximal sum of money in bourles can Vasya get by the end of day n?

Input

The first line contains two integers n and b (1 ≤ n, b ≤ 2000) — the number of days and the initial number of money in bourles. The next line contains n integers ai (1 ≤ ai ≤ 2000) — the prices of Martian dollars.

Output

Print the single number — which maximal sum of money in bourles can Vasya get by the end of day n.

Examples

Input

2 4
3 7


Output

8


Input

4 10
4 3 2 1


Output

10


Input

4 10
4 2 3 1


Output

15
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.StringTokenizer;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskB {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n = in.nextInt();
        int b = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }
        int[] max = new int[n];
        for (int i = n - 1; i >= 0; i--) {
            if (i + 1 < n) {
                max[i] = Math.max(a[i], max[i + 1]);
            } else {
                max[i] = a[i];
            }
        }
        int res = b;
        for (int i = 0; i < n; i++) {
            int canBuy = b / a[i];
            int profit = (max[i] - a[i]) * canBuy;
            res = Math.max(res, b + profit);
        }
        out.println(res);
    }
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

}
