Problem: You are given an unweighted tree with n vertices. Then n - 1 following operations are applied to the tree. A single operation consists of the following steps: 

  1. choose two leaves; 
  2. add the length of the simple path between them to the answer; 
  3. remove one of the chosen leaves from the tree. 



Initial answer (before applying operations) is 0. Obviously after n - 1 such operations the tree will consist of a single vertex. 

Calculate the maximal possible answer you can achieve, and construct a sequence of operations that allows you to achieve this answer!

Input

The first line contains one integer number n (2 ≤ n ≤ 2·105) — the number of vertices in the tree. 

Next n - 1 lines describe the edges of the tree in form ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi). It is guaranteed that given graph is a tree.

Output

In the first line print one integer number — maximal possible answer. 

In the next n - 1 lines print the operations in order of their applying in format ai, bi, ci, where ai, bi — pair of the leaves that are chosen in the current operation (1 ≤ ai, bi ≤ n), ci (1 ≤ ci ≤ n, ci = ai or ci = bi) — choosen leaf that is removed from the tree in the current operation. 

See the examples for better understanding.

Examples

Input

3
1 2
1 3


Output

3
2 3 3
2 1 1


Input

5
1 2
1 3
2 4
2 5


Output

9
3 5 5
4 3 3
4 1 1
4 2 2

Solution (in PYTHON3):

Solution:
import sys

n = int(sys.stdin.buffer.readline().decode('utf-8'))
adj = [[] for _ in range(n)]
deg = [0]*n
for u, v in (map(int, line.decode('utf-8').split()) for line in sys.stdin.buffer):
    adj[u-1].append(v-1)
    adj[v-1].append(u-1)
    deg[u-1] += 1
    deg[v-1] += 1


def get_dia(adj):
    from collections import deque
    n = len(adj)
    dq = deque([(0, -1)])

    while dq:
        end1, par = dq.popleft()
        for dest in adj[end1]:
            if dest != par:
                dq.append((dest, end1))

    prev = [-1]*n
    prev[end1] = -2
    dq = deque([(end1, 0)])

    while dq:
        end2, diameter = dq.popleft()
        for dest in adj[end2]:
            if prev[dest] == -1:
                prev[dest] = end2
                dq.append((dest, diameter+1))

    return end1, end2, diameter, prev


def get_dist(st, dist):
    dist[st] = 0
    stack = [st]
    while stack:
        v = stack.pop()
        for dest in adj[v]:
            if dist[dest] != -1:
                continue
            dist[dest] = dist[v] + 1
            stack.append(dest)


end1, end2, _, prev = get_dia(adj)
dia_path = [0]*n

x = end2
while x != end1:
    dia_path[x] = 1
    x = prev[x]
dia_path[x] = 1

dist1, dist2 = [-1]*n, [-1]*n
get_dist(end1, dist1)
get_dist(end2, dist2)
ans = []
score = 0
stack = [i for i in range(n) if deg[i] == 1 and dia_path[i] == 0]

while stack:
    v = stack.pop()
    deg[v] = 0
    if dist1[v] > dist2[v]:
        score += dist1[v]
        ans.append(f'{end1+1} {v+1} {v+1}')
    else:
        score += dist2[v]
        ans.append(f'{end2+1} {v+1} {v+1}')
    for dest in adj[v]:
        if dia_path[dest] == 0 and deg[dest] > 0:
            deg[dest] -= 1
            if deg[dest] <= 1:
                stack.append(dest)

x = end2
while x != end1:
    score += dist1[x]
    ans.append(f'{end1+1} {x+1} {x+1}')
    x = prev[x]

sys.stdout.buffer.write(
    (str(score) + '\n' + '\n'.join(ans)).encode('utf-8')
)