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.

Incorrect submission (verdict: Wrong Answer):
import sys
input = sys.stdin.buffer.readline
from collections import deque

def main():
    N = int(input())
    edge = [[] for _ in range(N)]
    for _ in range(N-1):
        a,b = map(int,input().split())
        edge[a-1].append(b-1)
        edge[b-1].append(a-1)
    
    for i in range(N):
        if len(edge[i]) == 1:
            start = i
            break

    rest = [[] for _ in range(3)]
    go = [False for _ in range(N)]
    rest[0].append(start)
    go[start] = True
    q = deque([start])
    use = 1
    while q:
        now = q.popleft()
        for fol in edge[now]:
            if not go[fol]:
                rest[use%3].append(fol)
                go[fol] = True
                q.append(fol)
        use += 1
    
    t = rest[0]+rest[1]+rest[2]
    ans = [0 for _ in range(N)]
    for x,num in enumerate(t):
        ans[num] = x+1
        
    print(*ans)
    
if __name__ == "__main__":
    main()


Fixed, accepted submission:
import sys
input = sys.stdin.buffer.readline
from collections import deque

def main():
    N = int(input())
    edge = [[] for _ in range(N)]
    for _ in range(N-1):
        a,b = map(int,input().split())
        edge[a-1].append(b-1)
        edge[b-1].append(a-1)

    
    use = [-1 for _ in range(N)]
    d = [[0],[]]
    use[0] = 0
    q = deque([0])
    while q:
        now = q.popleft()
        for fol in edge[now]:
            if use[fol] == -1:
                use[fol] = (use[now]+1)%2
                q.append(fol)
                d[use[fol]].append(fol)

    ans = [0 for _ in range(N)]
    b = [False for _ in range(N)]
    if len(d[0]) <= N//3:
        l = len(d[0])
        for i in range(l):
            ans[d[0][i]] = 3*i+3
            b[3*i+2] = True
        t = 0
        for num in d[1]:
            while b[t]:
                t += 1
            ans[num] = t+1
            b[t] = True
    elif len(d[1]) <= N//3:
        l = len(d[1])
        for i in range(l):
            ans[d[1][i]] = 3*i+3
            b[3*i+2] = True
        t = 0
        for num in d[0]:
            while b[t]:
                t += 1
            ans[num] = t+1
            b[t] = True
    else:
        l = len(d[0])
        t = 0
        for i in range(l):
            if i < (N-1)//3+1:
                ans[d[0][i]] = 3*i+1
                b[3*i] = True
            else:
                ans[d[0][i]] = 3*t+3
                b[3*t+2] = True
                t += 1
        l = len(d[1])
        for i in range(l):
            if i < (N-2)//3+1:
                ans[d[1][i]] = 3*i+2
                b[3*i+1] = True
            else:
                ans[d[1][i]] = 3*t+3
                b[3*t+2] = True
                t += 1
    
    print(*ans)

if __name__ == "__main__":
    main()