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.

Examples

Input

3 100
1 2
2 3


Output

3
4
3


Input

4 100
1 2
1 3
1 4


Output

8
5
5
5


Input

1 100


Output

1


Input

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


Output

0
0
1
1
1
0
1
0
1
1
# -*- 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)