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: Time Limit Exceeded)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Console;
using System.Text;
using System.IO;
using static System.Math;
namespace AtCoder
{
    class Program
    {
        static public long[] Sarray() { return ReadLine().Split().Select(long.Parse).ToArray(); }
        static public List<long> Slist() { return ReadLine().Split().Select(long.Parse).ToList(); }
        
        static readonly long Mod = 1000000000 + 7;
        static void Main(string[] args)
        {
            var NM = Sarray();
            N = NM[0];
            M = NM[1];
            a = new int[N].Select(_ => new List<long>()).ToArray();
            for (var i = 0; i < N - 1; ++i)
            {
                var t = Sarray();
                a[t[0] - 1].Add(t[1] - 1);
                a[t[1] - 1].Add(t[0] - 1);
            }
            dpc = new int[N].Select(_ => 1L).ToArray();
            dp2 = new int[N].Select(_ => new List<long>()).ToArray();
            parent = new long[N];

            //ある頂点を親としたとき、子の取る数を求めていく(dp2)
            //子が葉なら2(自分が黒のとき、子は白、黒の2通り)
            //子の子･･･とあるなら子が取る数+1(子が黒の時、子が白の時)

            dfs(0, -1);
            dpa = new int[N].Select(_ => 1L).ToArray();
            dfs2(0, -1);
            for (var i = 0; i < N; ++i)
                WriteLine(dpa[i]);
        }
        static List<long>[] a;
        static long N;
        static long M;
        static long[] dpc;
        static long[] dpa;
        static List<long>[] dp2;
        static long[] parent;
        static void dfs(long v, long p)
        {
            var cnt = 0;
            parent[v] = p;
            foreach(var n in a[v])
            {
                if (n == p)
                {
                    dp2[v].Add(0);
                    continue;
                }
                cnt++;
                dfs(n, v);
                dp2[v].Add(dpc[n] + 1);
                dpc[v] = dpc[v] * (dpc[n] + 1) % M;
            }
        }
        static void dfs2(long v, long p)
        {
            for (var j = 0; j < a[v].Count; ++j) 
            {
                //親について求める
                var n = a[v][j];
                if (n == p)
                {
                    long t = 1;
                    for (var i = 0; i < a[p].Count; ++i)
                    {
                        if (a[p][i] == v) continue;//自分への枝は無視
                        t = t * dp2[p][i] % M;
                    }
                    dp2[v][j] = (t + 1) % M;
                }
                //伸びる枝に対して
                dpa[v] = dpa[v] * dp2[v][j] % M;
            }
            foreach (var n in a[v])
            {
                if (n == p) continue;
                dfs2(n, v);
            }
        }
    }
}