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 Python (3.8.2) (judge verdict: Accepted)

from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
from bisect import bisect_right, bisect_left
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor, gamma, log
from operator import mul
from functools import reduce
from copy import deepcopy

sys.setrecursionlimit(2147483647)
INF = 10 ** 20
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): pass
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]


n, mod = LI()
G = [[] for _ in range(n)]
for _ in range(n - 1):
    a, b = LI()
    G[a - 1] += [b - 1]
    G[b - 1] += [a - 1]

dp = [-1] * n
par = [-1] * n
def f(u):
    ret = 1
    for v in G[u]:
        if v == par[u]:
            continue
        par[v] = u
        ret = ret * f(v)
    dp[u] = ret
    if u: return ret + 1

f(0)
ans = [0] * n
ans[0] = dp[0]
par_acc = [1] * n
dp2 = [1] * n
q = deque([0])
while q:
    u = q.pop()
    ans[u] = dp[u] * dp2[u] % mod
    L1 = [1]
    L2 = []
    D = {}
    cnt = 0
    for v in G[u]:
        if par[u] == v:
            continue
        L1 += [dp[v] + 1]
        L2 += [dp[v] + 1]
        D[cnt] = v
        q += [v]
        cnt += 1
    L2 += [1]
    if len(L1) > 2:
        for i in range(len(L1) - 2):
            L1[i + 2] = L1[i + 2] * L1[i + 1] % mod
            L2[-i - 3] = L2[-i - 3] * L2[-i - 2] % mod
    for j in D:
        dp2[D[j]] = (L1[j] * L2[j + 1] * dp2[u] + 1) % mod

print(*ans, sep="\n")