Problem: This problem is split into two tasks. In this task, you are required to find the minimum possible answer. In the task Village (Maximum) you are required to find the maximum possible answer. Each task is worth 50 points.

There are N houses in a certain village. A single villager lives in each of the houses. The houses are connected by roads. Each road connects two houses and is exactly 1 kilometer long. From each house it is possible to reach any other using one or several consecutive roads. In total there are N-1 roads in the village.

One day all villagers decided to move to different houses — that is, after moving each house should again have a single villager living in it, but no villager should be living in the same house as before. We would like to know the smallest possible total length in kilometers of the shortest paths between the old and the new houses for all villagers.

<image>

Example village with seven houses 

For example, if there are seven houses connected by roads as shown on the figure, the smallest total length is 8 km (this can be achieved by moving 1 → 6, 2 → 4, 3 → 1, 4 → 2, 5 → 7, 6 → 3, 7 → 5).

Write a program that finds the smallest total length of the shortest paths in kilometers and an example assignment of the new houses to the villagers.

Input

The first line contains an integer N (1 < N ≤ 10^5). Houses are numbered by consecutive integers 1, 2, …, N.

Then N-1 lines follow that describe the roads. Each line contains two integers a and b (1 ≤ a, b ≤ N, a ≠ b) denoting that there is a road connecting houses a and b.

Output

In the first line output the smallest total length of the shortest paths in kilometers.

In the second line describe one valid assignment of the new houses with the smallest total length: N space-separated distinct integers v_1, v_2, …, v_N. For each i, v_i is the house number where the villager from the house i should move (v_i ≠ i). If there are several valid assignments, output any of those.

Scoring

Subtasks: 

  1. (6 points) N ≤ 10 
  2. (19 points) N ≤ 1 000 
  3. (25 points) No further constraints 

Examples

Input


4
1 2
2 3
3 4


Output


4
2 1 4 3


Input


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


Output


8
3 4 6 2 7 1 5

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const unsigned int maxn = 1e5;
struct Edge {
  unsigned int to;
  const Edge* pre;
} ed[maxn * 2 + 10];
const Edge* head[maxn + 10];
unsigned int n;
inline void addEdge(const unsigned int from, const unsigned int to) {
  static Edge* cur = ed;
  new (cur) Edge{to, head[from]};
  head[from] = cur++;
  new (cur) Edge{from, head[to]};
  head[to] = cur++;
}
unsigned int dfsCentroid(const unsigned int x, const unsigned int f,
                         unsigned int& pos, unsigned int& mxs) {
  unsigned int ret = 1, ms = 0;
  for (const Edge* i = head[x]; i; i = i->pre) {
    const unsigned int to = i->to;
    if (to == f) continue;
    const unsigned int sz = dfsCentroid(to, x, pos, mxs);
    ret += sz;
    ms = max(ms, sz);
  }
  ms = max(ms, n - ret);
  if (ms < mxs) {
    mxs = ms;
    pos = x;
  }
  return ret;
}
unsigned int treeCentroid(const unsigned int root, const unsigned int n) {
  unsigned int ret, ms = UINT_MAX;
  dfsCentroid(root, 0, ret, ms);
  return ret;
}
namespace MinDist {
unsigned int seq[maxn + 10];
template <bool isRoot>
unsigned int dfs(const unsigned int x, const unsigned int f) {
  unsigned int ret = 0;
  for (const Edge* i = head[x]; i; i = i->pre)
    if (i->to != f) ret += dfs<false>(i->to, x);
  if (seq[x] == x) {
    ret += 2;
    if constexpr (isRoot)
      swap(seq[x], seq[head[x]->to]);
    else
      swap(seq[x], seq[f]);
  }
  return ret;
}
static unsigned int solve(const unsigned int n) {
  iota(seq + 1, seq + 1 + n, 1);
  return dfs<true>(1, 0);
}
}  // namespace MinDist
namespace MaxDist {
unsigned int dfn[maxn + 10], rev[maxn + 10], siz[maxn + 10];
unsigned int seq[maxn + 10];
unsigned long long dfs(const unsigned int x, const unsigned int f) {
  static unsigned int sta = 0;
  dfn[x] = sta++;
  rev[dfn[x]] = x;
  siz[x] = 1;
  unsigned long long ret = 0;
  for (const Edge* i = head[x]; i; i = i->pre) {
    const unsigned int to = i->to;
    if (to == f) continue;
    ret += dfs(to, x) + 2 * min(n - siz[to], siz[to]);
    siz[x] += siz[to];
  }
  return ret;
}
unsigned int maxSubtree(const unsigned int x) {
  unsigned int ret = 0;
  for (const Edge* i = head[x]; i; i = i->pre) ret = max(ret, siz[i->to]);
  return ret;
}
unsigned long long solve(const unsigned int n) {
  const unsigned int rt = treeCentroid(1, n);
  const unsigned long long ret = dfs(rt, 0);
  const unsigned int ms = maxSubtree(rt);
  for (unsigned int i = 1; i <= n; ++i) seq[i] = rev[(dfn[i] + ms) % n];
  return ret;
}
}  // namespace MaxDist
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (unsigned int i = 1; i < n; ++i) {
    unsigned int a, b;
    cin >> a >> b;
    addEdge(a, b);
  }
  cout << MaxDist::solve(n) << "\n";
  copy(MaxDist::seq + 1, MaxDist::seq + 1 + n,
       ostream_iterator<unsigned int>(cout, " "));
  cout.put('\n');
  return 0;
}