Problem:
Score : 300 points

Problem Statement

We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.

Initially, the hand of every clock stands still, pointing directly upward.

Now, Dolphin starts all the clocks simultaneously.

In how many seconds will the hand of every clock point directly upward again?

Constraints

1≤N≤100

1≤T_i≤10^{18}

All input values are integers.

The correct answer is at most 10^{18} seconds.

Input

Input is given from Standard Input in the following format:

N
T_1
:
T_N

Output

Print the number of seconds after which the hand of every clock point directly upward again.

Sample Input 1

2
2
3

Sample Output 1

6

We have two clocks. The time when the hand of each clock points upward is as follows:

Clock 1: 2, 4, 6, ... seconds after the beginning

Clock 2: 3, 6, 9, ... seconds after the beginning

Therefore, it takes 6 seconds until the hands of both clocks point directly upward.

Sample Input 2

5
2
5
10
1000000000000000000
1000000000000000000

Sample Output 2

1000000000000000000

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.Closeable;
import java.io.IOException;
import java.util.NoSuchElementException;
import java.io.InputStream;

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

    static class CMultipleClocks {
        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            int N = in.nextInt();
            long[] T = in.nextLongArray(N);
            long ans = T[0];
            for (int i = 1; i < N; i++) {
                long gcd = gcd(ans, T[i]);
                ans = T[i] * (ans / gcd);
            }
            out.println(ans);
        }

        private static long gcd(long m, long n) {
            if (m < n) return gcd(n, m);
            if (n == 0) return m;
            return gcd(n, m % n);
        }

    }

    static class FastScanner implements Closeable {
        private final InputStream in;
        private final byte[] buffer = new byte[1024];
        private int ptr = 0;
        private int buflen = 0;

        public FastScanner(InputStream in) {
            this.in = in;
        }

        private boolean hasNextByte() {
            if (ptr < buflen) {
                return true;
            } else {
                ptr = 0;
                try {
                    buflen = in.read(buffer);
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (buflen <= 0) {
                    return false;
                }
            }
            return true;
        }

        private int readByte() {
            if (hasNextByte()) return buffer[ptr++];
            else return -1;
        }

        private static boolean isPrintableChar(int c) {
            return 33 <= c && c <= 126;
        }

        public boolean hasNext() {
            while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
            return hasNextByte();
        }

        public long nextLong() {
            if (!hasNext()) throw new NoSuchElementException();
            long n = 0;
            boolean minus = false;
            int b = readByte();
            if (b == '-') {
                minus = true;
                b = readByte();
            }
            if (b < '0' || '9' < b) {
                throw new NumberFormatException();
            }
            while (true) {
                if ('0' <= b && b <= '9') {
                    n *= 10;
                    n += b - '0';
                } else if (b == -1 || !isPrintableChar(b)) {
                    return minus ? -n : n;
                } else {
                    throw new NumberFormatException();
                }
                b = readByte();
            }
        }

        public int nextInt() {
            long nl = nextLong();
            if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
            return (int) nl;
        }

        public long[] nextLongArray(int n) {
            long[] res = new long[n];
            for (int i = 0; i < n; i++) {
                res[i] = nextLong();
            }
            return res;
        }

        public void close() {
            try {
                in.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }

    }
}