Problem: We have a tree with N vertices. The vertices are numbered 1 to N, and the i-th edge connects Vertex a_i and Vertex b_i.

Takahashi loves the number 3. He is seeking a permutation p_1, p_2, \ldots , p_N of integers from 1 to N satisfying the following condition:

* For every pair of vertices (i, j), if the distance between Vertex i and Vertex j is 3, the sum or product of p_i and p_j is a multiple of 3.



Here the distance between Vertex i and Vertex j is the number of edges contained in the shortest path from Vertex i to Vertex j.

Help Takahashi by finding a permutation that satisfies the condition.

Constraints

* 2\leq N\leq 2\times 10^5
* 1\leq a_i, b_i \leq N
* The given graph is a tree.

Input

Input is given from Standard Input in the following format:


N
a_1 b_1
a_2 b_2
\vdots
a_{N-1} b_{N-1}


Output

If no permutation satisfies the condition, print `-1`.

Otherwise, print a permutation satisfying the condition, with space in between. If there are multiple solutions, you can print any of them.

Example

Input

5
1 2
1 3
3 4
3 5


Output

1 2 5 4 3

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using PII = pair<int, int>;
using VI = vector<int>;

const int N = 2e5 + 5;
VI adj[N];
VI comp[2];
int ans[N];

void dfs(int u, int p, int d) {
  comp[d].push_back(u);
  for (int v: adj[u]) {
    if (v == p) continue;
    dfs(v, u, d ^ 1);
  }
}

int n;
int f[] = {3, 1, 2};
int get(int x) {
  if (f[x] > n) return -1;
  f[x] += 3;
  return f[x] - 3;
}
template<typename... Args>
int get(int x, Args&&... args) {
  int res = get(x);
  if (res == -1) res = get(args...);
  return res;
}

int main() {
#ifdef LOCAL_EXEC
//	freopen("sample.in", "r", stdin);
//	freopen("sample.out", "w", stdout);
#else
  ios_base::sync_with_stdio(false); cin.tie(NULL);
#endif

  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b; cin >> a >> b;
    --a, --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }

  dfs(0, 0, 0);


  if (comp[0].size() > comp[1].size()) {
    swap(comp[0], comp[1]);
  }

  if ((int) comp[0].size() <= n / 3) {
    for (int u: comp[0]) {
      ans[u] = get(0);
    }
    for (int v: comp[1]) {
      ans[v] = get(1, 2, 0);
    }
  } else {
    for (int u: comp[0]) {
      ans[u] = get(1, 0);
    }
    for (int v: comp[1]) {
      ans[v] = get(2, 0);
    }
  }

  for (int i = 0; i < n; i++) {
    cout << ans[i] << ' ';
  }
  cout << endl;

  return 0;
}