Problem:
Score : 100 points

Problem Statement

There is a tree with N vertices, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N - 1), the i-th edge connects Vertex x_i and y_i.

Taro has decided to paint each vertex in white or black, so that any black vertex can be reached from any other black vertex by passing through only black vertices.

You are given a positive integer M.
For each v (1 \leq v \leq N), answer the following question:

Assuming that Vertex v has to be black, find the number of ways in which the vertices can be painted, modulo M.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

2 \leq M \leq 10^9

1 \leq x_i, y_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N M
x_1 y_1
x_2 y_2
:
x_{N - 1} y_{N - 1}

Output

Print N lines.
The v-th (1 \leq v \leq N) line should contain the answer to the following question:

Assuming that Vertex v has to be black, find the number of ways in which the vertices can be painted, modulo M.

Sample Input 1

3 100
1 2
2 3

Sample Output 1

3
4
3

There are seven ways to paint the vertices, as shown in the figure below.
Among them, there are three ways such that Vertex 1 is black, four ways such that Vertex 2 is black and three ways such that Vertex 3 is black.

Sample Input 2

4 100
1 2
1 3
1 4

Sample Output 2

8
5
5
5

Sample Input 3

1 100

Sample Output 3

1

Sample Input 4

10 2
8 5
10 8
6 5
1 5
4 8
2 10
3 6
9 2
1 7

Sample Output 4

0
0
1
1
1
0
1
0
1
1

Be sure to print the answers modulo M.

Solution in C# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
using System.IO;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using System.Text;
using System.Numerics;
using System.Threading.Tasks;
using System.Text.RegularExpressions;
using static System.Math;
using Debug = System.Diagnostics.Debug;
using LayoutKind = System.Runtime.InteropServices.LayoutKind;
using MethodImplOptions = System.Runtime.CompilerServices.MethodImplOptions;
using MethodImplAttribute = System.Runtime.CompilerServices.MethodImplAttribute;


static class P
{
    public static int Mod;
    static void Main()
    {
        var nm = Console.ReadLine().Split().Select(int.Parse).ToArray();
        var n = nm[0];
        Mod = nm[1];
        List<int>[] neighbours = Enumerable.Repeat(0, n).Select(_ => new List<int>()).ToArray();
        int[][] edges = Enumerable.Repeat(0, n - 1).Select(_ => Console.ReadLine().Split().Select(x => int.Parse(x) - 1).ToArray()).ToArray();
        ReRooting treeDP = new ReRooting(n, edges);
        Console.WriteLine(string.Join("\n", Enumerable.Range(0, n).Select(x => treeDP.Query(x))));
    }
}


class ReRooting
{
    public int NodeCount { get; private set; }

    int[][] Neighbours;
    int[][] IndexForNeighbours;

    long[] Res;
    long[][] dp;
    public ReRooting(int nodeCount, int[][] edges)
    {
        NodeCount = nodeCount;
        List<int>[] neighbours = new List<int>[nodeCount];
        List<int>[] indexForNeighbours = new List<int>[nodeCount];

        for (int i = 0; i < nodeCount; i++)
        {
            neighbours[i] = new List<int>();
            indexForNeighbours[i] = new List<int>();
        }
        for (int i = 0; i < edges.Length; i++)
        {
            var edge = edges[i];
            indexForNeighbours[edge[0]].Add(neighbours[edge[1]].Count);
            indexForNeighbours[edge[1]].Add(neighbours[edge[0]].Count);
            neighbours[edge[0]].Add(edge[1]);
            neighbours[edge[1]].Add(edge[0]);
        }

        Neighbours = new int[nodeCount][];
        IndexForNeighbours = new int[nodeCount][];
        for (int i = 0; i < nodeCount; i++)
        {
            Neighbours[i] = neighbours[i].ToArray();
            IndexForNeighbours[i] = indexForNeighbours[i].ToArray();
        }

        dp = new long[Neighbours.Length][];
        Res = new long[Neighbours.Length];

        for (int i = 0; i < Neighbours.Length; i++) dp[i] = new long[Neighbours[i].Length];
        Initialize();
    }

    public long Query(int x) => Res[x];

    private void Initialize()
    {
        int[] parents = new int[NodeCount];
        int[] order = new int[NodeCount];

        #region InitOrderedTree
        var index = 0;
        Stack<int> stack = new Stack<int>();
        stack.Push(0);
        parents[0] = -1;
        while (stack.Count > 0)
        {
            var node = stack.Pop();
            order[index++] = node;
            for (int i = 0; i < Neighbours[node].Length; i++)
            {
                var neighbour = Neighbours[node][i];
                if (neighbour == parents[node]) continue;
                stack.Push(neighbour);
                parents[neighbour] = node;
            }
        }
        #endregion

        #region fromLeaf
        //DFSの逆順に辿ると、親に対して値を返すために必要な情報は既に集まっている。
        for (int i = order.Length - 1; i >= 1; i--)
        {
            var node = order[i];
            var parent = parents[node];

            long accum = 1;
            int neighbourIndex = -1;
            for (int j = 0; j < Neighbours[node].Length; j++)
            {
                if (Neighbours[node][j] == parent)
                {
                    neighbourIndex = j;
                    continue;
                }
                //部分木の結果を累積
                accum = (accum * dp[node][j]) % P.Mod;
            }
            //自ノードの値を作用させ、dpに保持させる
            dp[parent][IndexForNeighbours[node][neighbourIndex]] = accum + 1;
        }
        #endregion

        #region toLeaf
        //既にノードの値を出すために必要な情報は集まっているので、出す
        for (int i = 0; i < order.Length; i++)
        {
            var node = order[i];

            long accum = 1;

            //尻からの累積
            long[] accumsFromTail = new long[Neighbours[node].Length];
            accumsFromTail[accumsFromTail.Length - 1] = 1;
            for (int j = accumsFromTail.Length - 1; j >= 1; j--) accumsFromTail[j - 1] = (accumsFromTail[j] * dp[node][j]) % P.Mod;

            //頭から累積をしながら配る
            for (int j = 0; j < accumsFromTail.Length; j++)
            {
                dp[Neighbours[node][j]][IndexForNeighbours[node][j]] = (accum * accumsFromTail[j]) % P.Mod + 1;
                accum = (accum * dp[node][j]) % P.Mod;
            }
            //累積し終えたものにノードの値を作用させたものが結果(今回はなし)
            Res[node] = accum;
        }
        #endregion
    }
}