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:
from sys import stdin, stdout
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
        
mod = pow(10, 9) + 7
mod2 = 998244353
        
def inp(): return stdin.readline().strip()
def iinp(): return int(inp())
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def remadd(x, y): return 1 if x%y else 0
def ceil(a,b): return (a+b-1)//b
  
def isprime(x):
    if x<=1: return False
    if x in (2, 3): return True
    if x%2 == 0: return False
    for i in range(3, int(sqrt(x))+1, 2):
        if x%i == 0: return False
    return True
 
n = iinp()
pars = [i for i in range(n+1)]
to_close = []
to_open = []
for i in range(n-1):
    x, y = mp()
    if pars[x] == pars[y]:
        to_close.append((x, y))
    else:
        for j in range(1, n+1):
            if y!=j and pars[j] == pars[y]:
                pars[j] = pars[x]
        pars[y] = pars[x]
s = set([pars[1]])
for i in range(2, n+1):
    if pars[i] not in s:
        to_open.append((1, i))
        s.add(pars[i])
l = len(to_close)
print(l)
for i in range(l):
    print(to_close[i][0], to_close[i][1], to_open[i][0], to_open[i][1])