Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.

Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
The first line contains integer *q* (1 ≤ *q* ≤ 1000), the number of handle change requests.

Next *q* lines contain the descriptions of the requests, one per line.

Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.

The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone.
# -*- coding: utf-8 -*-
# Link al problema: https://codeforces.com/problemset/problem/501/B
# Usamos una tabla de hash en la que se mantienen los nombres originales con
# una referencia al más reciente y los más recientes con referencia la original
import sys


if __name__ == "__main__":
    handles = sys.stdin.readlines()
    handles = [handle.rstrip('\n').split(" ") for handle in handles[1:]]
    users = {}
    originals = []

    for old, new in handles:
        if old not in users.keys():
            # Agregamos un nombre original
            users[old] = new
            originals.append(old)
            # Agregamos su primer cambio
            users[new] = old
        else:
            previous = old
            old = users[old]
            # Como tendremos referencia directa no necesitamos a este
            del users[previous]

            # El nuevo nombre tiene referencia directa al original
            users[new] = old
            # Aumentamos contador de cambios de nombre original
            users[old] = new

    sys.stdout.write("{}\n".format(len(originals)))
    for handle in originals:
        sys.stdout.write("{} {}\n".format(handle, users[handle]))
