output_description: 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.
  
  * * *
input_description: 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}
attempt: # -*- coding: utf-8 -*-
  
  #############
  # Libraries #
  #############
  
  import sys
  
  input = sys.stdin.readline
  
  import math
  
  # from math import gcd
  import bisect
  import heapq
  from collections import defaultdict
  from collections import deque
  from collections import Counter
  from functools import lru_cache
  
  #############
  # Constants #
  #############
  
  
  INF = float("inf")
  AZ = "abcdefghijklmnopqrstuvwxyz"
  
  #############
  # Functions #
  #############
  
  
  ######INPUT######
  def I():
      return int(input().strip())
  
  
  def S():
      return input().strip()
  
  
  def IL():
      return list(map(int, input().split()))
  
  
  def SL():
      return list(map(str, input().split()))
  
  
  def ILs(n):
      return list(int(input()) for _ in range(n))
  
  
  def SLs(n):
      return list(input().strip() for _ in range(n))
  
  
  def ILL(n):
      return [list(map(int, input().split())) for _ in range(n)]
  
  
  def SLL(n):
      return [list(map(str, input().split())) for _ in range(n)]
  
  
  N, MOD = IL()
  
  
  #####Shorten#####
  def DD(arg):
      return defaultdict(arg)
  
  
  #####Inverse#####
  def inv(n):
      return pow(n, MOD - 2, MOD)
  
  
  ######Combination######
  kaijo_memo = []
  
  
  def kaijo(n):
      if len(kaijo_memo) > n:
          return kaijo_memo[n]
      if len(kaijo_memo) == 0:
          kaijo_memo.append(1)
      while len(kaijo_memo) <= n:
          kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
      return kaijo_memo[n]
  
  
  gyaku_kaijo_memo = []
  
  
  def gyaku_kaijo(n):
      if len(gyaku_kaijo_memo) > n:
          return gyaku_kaijo_memo[n]
      if len(gyaku_kaijo_memo) == 0:
          gyaku_kaijo_memo.append(1)
      while len(gyaku_kaijo_memo) <= n:
          gyaku_kaijo_memo.append(
              gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo), MOD - 2, MOD) % MOD
          )
      return gyaku_kaijo_memo[n]
  
  
  def nCr(n, r):
      if n == r:
          return 1
      if n < r or r < 0:
          return 0
      ret = 1
      ret = ret * kaijo(n) % MOD
      ret = ret * gyaku_kaijo(r) % MOD
      ret = ret * gyaku_kaijo(n - r) % MOD
      return ret
  
  
  ######Factorization######
  def factorization(n):
      arr = []
      temp = n
      for i in range(2, int(-(-(n**0.5) // 1)) + 1):
          if temp % i == 0:
              cnt = 0
              while temp % i == 0:
                  cnt += 1
                  temp //= i
              arr.append([i, cnt])
      if temp != 1:
          arr.append([temp, 1])
      if arr == []:
          arr.append([n, 1])
      return arr
  
  
  #####MakeDivisors######
  def make_divisors(n):
      divisors = []
      for i in range(1, int(n**0.5) + 1):
          if n % i == 0:
              divisors.append(i)
              if i != n // i:
                  divisors.append(n // i)
      return divisors
  
  
  #####MakePrimes######
  def make_primes(N):
      max = int(math.sqrt(N))
      seachList = [i for i in range(2, N + 1)]
      primeNum = []
      while seachList[0] <= max:
          primeNum.append(seachList[0])
          tmp = seachList[0]
          seachList = [i for i in seachList if i % tmp != 0]
      primeNum.extend(seachList)
      return primeNum
  
  
  #####GCD#####
  def gcd(a, b):
      while b:
          a, b = b, a % b
      return a
  
  
  #####LCM#####
  def lcm(a, b):
      return a * b // gcd(a, b)
  
  
  #####BitCount#####
  def count_bit(n):
      count = 0
      while n:
          n &= n - 1
          count += 1
      return count
  
  
  #####ChangeBase#####
  def base_10_to_n(X, n):
      if X // n:
          return base_10_to_n(X // n, n) + [X % n]
      return [X % n]
  
  
  def base_n_to_10(X, n):
      return sum(int(str(X)[-i - 1]) * n**i for i in range(len(str(X))))
  
  
  def base_10_to_n_without_0(X, n):
      X -= 1
      if X // n:
          return base_10_to_n_without_0(X // n, n) + [X % n]
      return [X % n]
  
  
  #####IntLog#####
  def int_log(n, a):
      count = 0
      while n >= a:
          n //= a
          count += 1
      return count
  
  
  #############
  # Main Code #
  #############
  
  
  order = []
  d = [INF] * N
  
  
  def bfs(graph, start):
      N = len(graph)
      d[start] = 0
      q = deque([start])
      while q:
          u = q.popleft()
          order.append(u)
          for v in graph[u]:
              if d[v] != INF:
                  continue
              d[v] = d[u] + 1
              q.append(v)
  
  
  graph = [[] for i in range(N)]
  
  for _ in range(N - 1):
      a, b = IL()
      graph[a - 1].append(b - 1)
      graph[b - 1].append(a - 1)
  
  bfs(graph, 0)
  
  dp = [0 for i in range(N)]
  cum = [None for i in range(N)]
  for x in order[::-1]:
      cum1 = [1]
      for y in graph[x]:
          if d[x] < d[y]:
              cum1.append((cum1[-1] * (dp[y] + 1)) % MOD)
      cum2 = [1]
      for y in graph[x][::-1]:
          if d[x] < d[y]:
              cum2.append((cum2[-1] * (dp[y] + 1)) % MOD)
      dp[x] = cum1[-1]
      cum[x] = [cum1, cum2]
  
  factor = [1 for i in range(N)]
  for x in order:
      i = 0
      for y in graph[x]:
          if d[x] < d[y]:
              dpinved = cum[x][0][i] * cum[x][1][len(cum[x][0]) - 2 - i]
              factor[y] = (dpinved * factor[x] + 1) % MOD
              i += 1
  
  for i in range(N):
      print((dp[i] * factor[i]) % MOD)
  
problem_description: 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.
samples: [{"input": "3 100\n    1 2\n    2 3", "output": "3\n    4\n    3\n    \n\nThere are seven ways to paint the vertices, as shown in the figure below.\nAmong them, there are three ways such that Vertex 1 is black, four ways such\nthat Vertex 2 is black and three ways such that Vertex 3 is black.\n\n![](https://img.atcoder.jp/dp/subtree_0_muffet.png)\n\n* * *"}, {"input": "4 100\n    1 2\n    1 3\n    1 4", "output": "8\n    5\n    5\n    5\n    \n\n* * *"}, {"input": "1 100", "output": "1\n    \n\n* * *"}, {"input": "10 2\n    8 5\n    10 8\n    6 5\n    1 5\n    4 8\n    2 10\n    3 6\n    9 2\n    1 7", "output": "0\n    0\n    1\n    1\n    1\n    0\n    1\n    0\n    1\n    1\n    \n\nBe sure to print the answers modulo M."}]