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 org.omg.PortableInterceptor.SYSTEM_EXCEPTION;

import java.io.*;
import java.util.*;
import java.util.regex.Matcher;

public class Main {

    public static void main(String[] args) throws IOException {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = /*new InputReader(new FileReader("input.txt"));*/   new InputReader(inputStream);
        PrintWriter out = /*new PrintWriter("output.txt"); */  new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(in, out);
        out.close();
    }

    private static class TaskB {

        static final long max = 1000000000000000000L;
        static final double eps = 0.000000001;
        static final long mod = 1000000007;

        void solve(InputReader in, PrintWriter out) throws IOException {

            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 maxi = B;
            for (int i = 0; i < N - 1; i++) {

                int j = i;
                int k = j + 1;
                int sum = B;

                while (k < N) {
                    if (A[k] > A[j]) {
                        sum = (sum / A[j]) * A[k] + sum % A[j];
                        j = k;
                        k = j + 1;
                    } else k++;
                }

                if (sum > maxi) maxi = sum;

            }
            out.println(maxi);
        }
    }

    private static class InputReader {
        StringTokenizer st;
        BufferedReader br;

        public InputReader(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public InputReader(FileReader s) throws FileNotFoundException {
            br = new BufferedReader(s);
        }

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

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

        public long nextLong() {
            return Long.parseLong(next());
        }

        public String nextLine() {
            try {
                return br.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public boolean ready() {
            try {
                return br.ready();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
    }
}