A + B is often used as an example of the easiest problem possible to show some contest platform. However, some scientists have observed that sometimes this problem is not so easy to get accepted. Want to try?

Input

The input contains two integers a and b (0 ≤ a, b ≤ 103), separated by a single space.

Output

Output the sum of the given integers.

Examples

Input

5 14


Output

19


Input

381 492


Output

873
//////////package codeforces;

import java.io.BufferedReader;
import java.io.Closeable;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.ListIterator;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class A implements Closeable {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter writer = new PrintWriter(System.out);

    A() throws IOException {
//        reader = new BufferedReader(new FileReader("input.txt"));
//        writer = new PrintWriter(new FileWriter("output.txt"));
    }

    StringTokenizer stringTokenizer;

    String next() throws IOException {
        while (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
            stringTokenizer = new StringTokenizer(reader.readLine());
        }
        return stringTokenizer.nextToken();
    }

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

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

    private final int MOD = 1000 * 1000 * 1000 + 7;

    int sum(int a, int b) {
        a += b;
        return a >= MOD ? a - MOD : a;
    }

    int product(int a, int b) {
        return (int) (1l * a * b % MOD);
    }

    int pow(int x, int k) {
        int result = 1;
        while (k > 0) {
            if (k % 2 == 1) {
                result = product(result, x);
            }
            x = product(x, x);
            k /= 2;
        }
        return result;
    }

    int inv(int x) {
        return pow(x, MOD - 2);
    }

    void solve() throws IOException {
        writer.println(nextInt() + nextInt());
    }

    public static void main(String[] args) throws IOException {
        try (A a = new A()) {
            a.solve();
        }
    }

    @Override
    public void close() throws IOException {
        reader.close();
        writer.close();
    }
}
