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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"
using namespace std;
#define FAST ios_base::sync_with_stdio(false); cin.tie(0);
#define LLINF (long long) 1e18//1234567890987654321
#define INF 1234567890
#define pb push_back
#define ins insert
#define f first
#define s second	
#define db 0
#define EPS (1e-7)    //0.0000001 the value
#define PI (acos(-1))
#define MAXN 300006
#define MAXK 26
#define MAXX 15000006
#define ll long long int
#define ld long double
#define rep0(kk, l1, l2)for(ll kk = l1; kk < l2; kk++)
#define rep1(kk, l1, l2)for(ll kk = l1; kk <= l2; kk++)
#define foritr(itr, A) for(set<ll>::iterator itr = A.begin(); itr != A.end(); itr++)
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());    //can be used by calling rng() or shuffle(A, A+n, rng)
#define FOR(ii, ss, ee) for(ll ii = ss; ii < ee; ii++)
#define cr(x) cerr << #x << " = " << x << "\n";
#define crA(x, A) cerr << #x << " = " << A[x] << "\n";
#define spacing if(db)cout << " ";
#define mmst(x, v) memset((x), v, sizeof ((x)));
#define bg(ms) (*ms.begin())
#define ed(ms) (*prev(ms.end(), 1))
#define addedge(a, b, c, v) v[(a)].pb(pi((b), (c))); v[(b)].pb(pi((a), (c)))
#define ph push
#define btinpct(x) __builtin_popcountll(x)
#define p2(x) (1LL<<(x))
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
typedef pair <ll, ll> pi;
typedef pair <ll, pi> spi;
typedef pair <pi, pi> dpi;
inline ll rand(ll x, ll y) { ++y; return (rng() % (y-x)) + x; } //inclusivesss
ll n, m, dp[MAXN], reall[MAXN], ans[MAXN];
vector <ll> v[MAXN];
void dfs(ll x, ll p)
{
	dp[x] = 1;
	for(auto i : v[x])
	{
		if(i==p) continue;
		dfs(i, x);
		dp[x] = (dp[x] * (dp[i]+1))%m;
	}
	// cout << x << ' ' << dp[x] << '\n';
	return;
}
ll divi(ll x, ll y)
{
	while(x < y) { x += m; }
	return x/y;
}
void real_calc(ll x, ll p)
{
	reall[x] = 1;
	for(auto i : v[x])
	{
		if(i==p) { reall[x] = (reall[x] * (divi(reall[i], (dp[x]+1))%m + 1)%m)%m; continue; }
		reall[x] = (reall[x] * (dp[i]+1))%m;
	}
	for(auto i : v[x])
	{
		if(i==p) continue;
		real_calc(i, x); 
	}
	return;
}
int main()
{
	cin >> n >> m; // m = LLINF;
	FOR(i,1,n)
	{
		ll a, b;
		cin >> a >> b; 
		// a = i, b = a+1; // cout << a << ' ' << b << '\n';
		v[a].pb(b);
		v[b].pb(a);
	}
	// cout << n << "\n";
	dfs(1, 1);
	real_calc(1, 1);
	FOR(i,1,n+1) cout << reall[i]%m << '\n';
}