Problem: Berland Government decided to improve relations with neighboring countries. First of all, it was decided to build new roads so that from each city of Berland and neighboring countries it became possible to reach all the others. There are n cities in Berland and neighboring countries in total and exactly n - 1 two-way roads. Because of the recent financial crisis, the Berland Government is strongly pressed for money, so to build a new road it has to close some of the existing ones. Every day it is possible to close one existing road and immediately build a new one. Your task is to determine how many days would be needed to rebuild roads so that from each city it became possible to reach all the others, and to draw a plan of closure of old roads and building of new ones.

Input

The first line contains integer n (2 ≤ n ≤ 1000) — amount of cities in Berland and neighboring countries. Next n - 1 lines contain the description of roads. Each road is described by two space-separated integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — pair of cities, which the road connects. It can't be more than one road between a pair of cities. No road connects the city with itself.

Output

Output the answer, number t — what is the least amount of days needed to rebuild roads so that from each city it became possible to reach all the others. Then output t lines — the plan of closure of old roads and building of new ones. Each line should describe one day in the format i j u v — it means that road between cities i and j became closed and a new road between cities u and v is built. Cities are numbered from 1. If the answer is not unique, output any.

Examples

Input

2
1 2


Output

0


Input

7
1 2
2 3
3 1
4 5
5 6
6 7


Output

1
3 1 3 7

Solution (in PYTHON3):

Solution:
import sys

lines = sys.stdin.read().split("\n")
lines.pop(-1)
n = int(lines.pop(0))
 
#Saque esta estructura de internet porque no habia otra
#implementacion nativa en python y no queria calentarme
class Node:
    """Represents an element of a set."""
    def __init__(self, id):
        self.id = id
        self.parent = self
        self.rank = 0
        self.size = 1
 
    def __repr__(self):
        return 'Node({!r})'.format(self.id)
 
 
def Find(x):
    """Returns the representative object of the set containing x."""
    if x.parent is not x:
        x.parent = Find(x.parent)
    return x.parent
 
 
def Union(x, y):
    """Combines the sets x and y belong to."""
    xroot = Find(x)
    yroot = Find(y)
 
    # x and y are already in the same set
    if xroot is yroot:
        return
 
    # x and y are not in same set, so we merge them
    if xroot.rank < yroot.rank:
        xroot, yroot = yroot, xroot  # swap xroot and yroot
 
    # merge yroot into xroot
    yroot.parent = xroot
    xroot.size += yroot.size
    if xroot.rank == yroot.rank:
        xroot.rank = xroot.rank + 1
 
nodos = [Node(i) for i in range(n+1)]
edges = []
final = []
for line in lines:
    a = nodos[int(line.split(" ")[0])]
    b = nodos[int(line.split(" ")[1])]
    if (Find(a) != Find(b)):
        Union(a, b)
    else:
        edges.append( (a,b) )
 
for i in range(2,n+1):
    if (Find(nodos[1]) != Find(nodos[i])):
        final.append( ( nodos[1],nodos[i] ) )
        Union(nodos[1],nodos[i])
        
print(len(final))
for i in range(len(final)):
    print(edges[i][0].id, end=" ")
    print(edges[i][1].id, end=" ")
    print(final[i][0].id, end=" ")
    print(final[i][1].id, end=" ")