911_F. Tree Destruction_16961
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
from sys import stdin, stdout
from itertools import repeat
def main():
    n = int(stdin.readline())
    dat = map(int, stdin.read().split(), repeat(10, 2 * n - 2))
    la = [None] * (n + 1)
    xt = [None] * len(dat)
    j = 0
    for i in xrange(n-1):
        x, y = dat[j], dat[j+1]
        xt[j] = la[y]
        la[y] = j
        j += 1
        xt[j] = la[x]
        la[x] = j
        j += 1
    st = [1]
    pu = st.append
    po = st.pop
    inf = n + 1
    d = [inf] * (n + 1)
    d[1] = 0
    while st:
        x = po()
        y = la[x]
        while y is not None:
            z = dat[y]
            if d[z] == inf:
                d[z] = d[x] + 1
                pu(z)
            y = xt[y]
    md = 0
    md1 = 1
    for i in xrange(1, n + 1):
        if md < d[i]:
            md = d[i]
            md1 = i
    d1 = [inf] * (n + 1)
    pu(md1)
    d = d1
    d[md1] = 0
    while st:
        x = po()
        y = la[x]
        while y is not None:
            z = dat[y]
            if d[z] == inf:
                d[z] = d[x] + 1
                pu(z)
            y = xt[y]
    md = 0
    md2 = 1
    for i in xrange(1, n + 1):
        if md < d[i]:
            md = d[i]
            md2 = i
    d2 = [inf] * (n + 1)
    pu(md2)
    d = d2
    d[md2] = 0
    while st:
        x = po()
        y = la[x]
        while y is not None:
            z = dat[y]
            if d[z] == inf:
                d[z] = d[x] + 1
                pu(z)
            y = xt[y]
    ans = []
    aa = md
    x = md1
    col = [0] * (n + 1)
    col[x] = 1
    for i in xrange(md):
        y = la[x]
        while y is not None:
            z = dat[y]
            if d[z] == d[x] - 1:
                col[z] = 1
                aa += d[z]
                ans.append((md1, z, z))
                x = z
                break
            y = xt[y]
    pu(md1)
    col[md1] = 2
    while st:
        x = po()
        y = la[x]
        while y is not None:
            z = dat[y]
            if col[z] != 2:
                if not col[z]:
                    if d1[z] > d2[z]:
                        aa += d1[z]
                        ans.append((md1, z, z))
                    else:
                        aa += d2[z]
                        ans.append((md2, z, z))
                col[z] = 2
                pu(z)
            y = xt[y]
    print aa
    stdout.write('\n'.join('%d %d %d' % x for x in reversed(ans)))
main()
