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;

const int MAXN = 200200;

int N;
vector<int> adj[MAXN];
int dist[MAXN];
vector<int> byDist[2];
bool wv[MAXN];
bool wa[MAXN];
int ans[MAXN];

void dfs(int v, int p) {
   byDist[dist[v] & 1].emplace_back(v);
   for (int u : adj[v]) {
      if (u != p) {
         dist[u] = dist[v] + 1;
         dfs(u, v);
      }
   }
}

int main() {
   ios_base::sync_with_stdio(false);
   cin >> N;
   for (int i = 0; i < N - 1; ++i) {
      int v, u;
      cin >> v >> u;
      adj[v].emplace_back(u);
      adj[u].emplace_back(v);
   }
   dfs(1, -1);
   if (byDist[0].size() > byDist[1].size()) {
      swap(byDist[0], byDist[1]);
   }
   if (byDist[0].size() <= (N / 3)) {
      int start = 3;
      for (int v : byDist[0]) {
         ans[v] = start;
         wv[v] = true;
         wa[start] = true;
         start += 3;
      }
   } else {
      int start = 2;
      for (int v : byDist[0]) {
         if (start <= N) {
            ans[v] = start;
            wv[v] = true;
            wa[start] = true;
            start += 3;
         }
      }
      start = 1;
      for (int v : byDist[1]) {
         if (start <= N) {
            ans[v] = start;
            wv[v] = true;
            wa[start] = true;
            start += 3;
         }
      }
   }
   int ptr = 1;
   for (int i = 1; i <= N; ++i) {
      if (!wv[i]) {
         while (wa[ptr]) ptr++;
         ans[i] = ptr++;
      }
      cout << ans[i] << " ";
   }
}