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>
#include <algorithm>
#include <cmath>
#include <set>
#include <cstdio>
#include <vector>
#include <iostream>
#include <utility>
#include <queue>
#include <map>
#include <complex>
#define fir first
#define sec second
#define sz(s) (s).size()
#define pb push_back
#define get(n) scanf("%d",&n);
#define gets(s) string s;cin >> (s);
#define prfi(n) printf("%d", &n);
#define prfd(n) printf("%lf", &n);
#define All(s) (s).begin(), (s).end()
#define rep(i,j) for(int (i)=0;(i)<(j);(i)++)
#define For(i,j,k) for(int (i)=(j);(i)<(k);(i)++)
#define drep(i,j) for(int (i)=(j);(i)>=0;(i)--)
#define Ford(i,j,k) for(int (i)=(j);i>=(k);i--)
#define fore(c,v) for(auto (c): (v))
#define lp for(int i=0;i<n;i++)
#define mem(a,b) memset(a,b,sizeof(a));
#define dump(x)  std::cout << #x << " = " << (x) << std::endl;
#define debug(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
using ull = unsigned long long int;
using ll = long long;
using ld = long double;
using pii = std::pair<int,int>;
using pll = std::pair<ll, ll>;
using vi = std::vector<int> ;
using vvi = std::vector<vi> ;
using vll = std::vector<ll>;
using vvll = std::vector<vll>;
using vd = std::vector<double> ;
using vvd = std::vector<vd> ;
using qi = std::queue<int> ;
using vpii = std::vector<std::pair<int, int> >;
using vpll = std::vector<pll>;
using namespace std;

const int Mod = (1e9) + 7;
const int INF = 1e9 + 19;
const ll INFL = 1e18 + 19;
const int dx[] = {-1, 0, 0, 1};
const int dy[] = {0, -1, 1, 0};
const int dx2[] = {-1, -1, -1, 0, 0, 0, 1, 1, 1};
const int dy2[] = {1, 0, -1, 1, 0, -1, 1, 0, -1};
const double EPS = 1e-10;
//_____________________________________Templates_________________________________________//

template<class T1, class T2> inline void chmin(T1 &a, T2 b){if(a > b) a = b;}
template<class T1, class T2> inline void chmax(T1 &a, T2 b){if(a < b) a = b;}
template<class T> inline void pri(T a){cout << a << endl;}
template<class Z> using vec = vector<Z>;
template<class T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
//mainly use for dynamic prog
template<class T1, class T2>
void update(T1 &a, T2 b){
  a += b;
  if(a > Mod) a %= Mod;
}

inline void IN(void){
  return;
}

template <typename First, typename... Rest>
void IN(First& first, Rest&... rest){
  cin >> first;
  IN(rest...);
  return;
}

inline void OUT(void){
  cout << "\n";
  return;
}

template <typename First, typename... Rest>
void OUT(First first, Rest... rest){
  cout << first << " ";
  OUT(rest...);
  return;
}

bool pairsort(pll pl, pll pr){
  if(pl.first == pr.first)return pl.second > pr.second;
  return pl.first < pr.first;
}

int cntbit(ll a,int n,int j){int res = 0;For(i,j,n){if(a>>i & 1){res++;}}return res;}
vector<int> make_bit(int a){vector<int> res; for(int i=31;i>=0;i--)if(a&(1<<i))res.pb(i);return res;}
bool stdbit(int a, int b){return a&(1 << b); }
int GCD(int a, int b){if(b > a)return GCD(b,a);if(a%b == 0)return b;else return GCD(b, a%b);}
int LCM(int a, int b){return a*b/GCD(a,b);}
int roundup(int a, int b){if(a % b == 0)return a/b;else return (a+b)/b;}
int rounddown(int a, int b){if(a%b == 0)return a/b;else {return (a-b)/b;}}
ll pow(ll a, ll n){ll res = 1;while(n > 0){if(n&1)res *= a; a *= a; n = n >> 1;}return res;}
ll GetDiviserCount(ll N)//約数の個数
{
  ll res = 1;
  For(i,2,sqrt(N)+1)
  {
    ll cnt = 0;
    while(N%i == 0)
    {
      cnt++;
      N /= i;
    }
    res *= (cnt + 1);
    if(N == 1)break;
  }
  if(N != 1)res *= 2;
  return res;
}
vll GetDivisor(ll N)//約数列挙
{
  vll res;
  for(ll i = 1;i*i <= N;i++)
  {
    if(N%i == 0)
    {
      res.pb(i);
      if(i*i != N)res.pb(N/i);
    }
  }
  sort(All(res));
  return res;
}


struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%Mod+Mod)%Mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= Mod) x -= Mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += Mod-a.x) >= Mod) x -= Mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= Mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
 
  // for prime mod
  mint inv() const {
    return pow(Mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
  friend ostream& operator<<(ostream& os, const mint& a);
};
ostream& operator<<(ostream& os, const mint& a)
{
  os << a.x;
  return os;
}
struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < Mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
};
 
template<class Z>
struct Matrix
{
  using mat = Matrix<Z>;
  vector<vector<Z>> m_dat;
  int m_h;
  int m_w;
  Matrix(int h, int w) : m_h(h), m_w(w) ,m_dat(h,vector<Z>(w)) {}
  vector<Z> &operator[] (int idx)
  {
    return m_dat[idx];
  }
  mat Multiple(mat &a)
  {
    mat C(m_h, a.m_w);
    for(int i=0;i<m_h;i++)
    {
      for(int j=0;j<a.m_w;j++)
      {
        for(int k=0;k<m_w;k++)
        {
         C[i][j] += m_dat[i][k] * a[k][j];
        }
      }
    }
    return C;
  }
  mat Pow(ll x)
  {
    mat B(m_h,m_w);
    for(int i=0;i<m_h;i++)
    {
      B[i][i] = 1;
    }
    mat A = *this;
    while(x > 0)
    {
      if(x&1)B = B.Multiple(A);
      A = A.Multiple(A);
      x = x >> 1;
    }
    return B;
  }
  friend ostream& operator<<(ostream &os, const mat &A);
};

template<class Z>
ostream& operator<<(ostream &os, Matrix<Z>& A)
{
  for(int i=0;i<A.m_h;i++)
  {
    for(int j=0;j<A.m_w;j++)
    {
      os << A[i][j] << " ";
    }
    os << endl;
  }
  return os;
}
//_____________________　following sorce code_________________________//
const int max_n = 3 * (1e5) + 1;
const int max_m = 83 * (1e5) + 1;

int n;
ll m;
ll N;
int h,w;
vvi tree;
string S;
int a,b,c;
vi v;
int ans;
double x,y,z;
vvll dp;

ll dfs(int v, int u=-1)
{
  ll res = 1;
  rep(i,tree[v].size())
  {
    int e = tree[v][i];
    if(e == u)continue;
    dp[v][i] = dfs(e,v)%m;
    (res *= (1+dp[v][i])%m) %= m;
  }
  
  return res%m;
}

void dfs2(int v, int u=-1,ll p = 0)
{
  if(u != -1)
  {
    rep(i,sz(tree[v]))
    {
      int e = tree[v][i];
      if(e == u)
      {
        dp[v][i] = p;
        break;
      }
    }
  }

  vll L(tree[v].size(),1);
  vll R(tree[v].size(),1);
  For(i,1,tree[v].size())
  {
    (L[i] = (L[i-1] *(1 + dp[v][i]))%m) %= m;
  }
  drep(i,tree[v].size()-1)
  {
    if(i != (tree[v].size()-1))(R[i] = (R[i+1] * (1 + dp[v][i+1]))%m)%m;
    ll tot = L[i]%m * R[i]%m;
    tot %= m;
    int e = tree[v][i];
    if(e == u)continue;
    dfs2(e,v,tot);
  }
}

signed main () {
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  IN(n,m);
  dp = vvll(n);
  tree = vvi(n);
  rep(i,n-1)
  {
    IN(a,b);a--;b--;
    tree[a].emplace_back(b);
    tree[b].emplace_back(a);
  }
  rep(i,n)
  {
    dp[i] = vll(tree[i].size());
  }
  dfs(0);
  dfs2(0);
  fore(&p, dp)
  {
    ll ans = 1;
    fore(e,p)
    {
      (ans *= (1 + e)%m) %= m;
    }
    cout << ans << endl;
  }
  //for(auto c : ans){cout << c << endl;}
  //cout << fixed << setprecision(15) << ans << endl;
  return 0;
}