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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
int MOD;

//辺の情報を個別に持つタイプ
typedef struct edge{ll s,g,c;}E;
typedef struct graph{
	int vcnt,ecnt;
	E  e[200010];//適宜変える(ecnt)
	int inv[200010];//逆辺のindex(ecnt)
	int id[100010];//適宜変える(vcnt)
}G;
G g;

int esort(const void*a,const void*b){
	E*p=(E*)a,*q=(E*)b;
	if((*p).s<(*q).s)return -1;
	if((*p).s>(*q).s)return  1;
	if((*p).g<(*q).g)return -1;
	return 1;
}
void makeinvedge();
void readgraph(){
	//適宜変える
	ll n;
	scanf("%lld%d",&n,&MOD);
	rep(i,0,n-1){
		ll x,y;
		scanf("%lld%lld",&x,&y);
		x--,y--;
		g.e[i].s=x;
		g.e[i].g=y;
	}
	g.vcnt=n;
	g.ecnt=n-1;
	qsort(g.e,g.ecnt,sizeof(E),esort);

	int p=0;
	rep(i,0,g.vcnt){
		while(p<g.ecnt&&g.e[p].s<i)p++;
		g.id[i]=p;
	}
	g.id[g.vcnt]=g.ecnt;//番兵
}

void makeinvedge(){
	//逆辺とidx
	int added=0;
	rep(i,0,g.ecnt){
		int l=0,r=g.ecnt;
		while(r-l>1){
			int m=(l+r)/2;
			if(g.e[m].s<g.e[i].g||(g.e[m].s==g.e[i].g&&g.e[m].g<=g.e[i].s))l=m;
			else r=m;
		}
		if(g.e[l].s!=g.e[i].g||g.e[l].g!=g.e[i].s){
			g.e[g.ecnt+added].s=g.e[i].g;
			g.e[g.ecnt+added].g=g.e[i].s;
			g.e[g.ecnt+added].c=0;
			added++;
		}
	}
	g.ecnt+=added;
	qsort(g.e,g.ecnt,sizeof(E),esort);

	int p=0;
	rep(i,0,g.vcnt){
		while(p<g.ecnt&&g.e[p].s<i)p++;
		g.id[i]=p;
	}
	g.id[g.vcnt]=g.ecnt;//番兵

	rep(i,0,g.ecnt){
		int l=0,r=g.ecnt;
		while(r-l>1){
			int m=(l+r)/2;
			if(g.e[m].s<g.e[i].g||(g.e[m].s==g.e[i].g&&g.e[m].g<=g.e[i].s))l=m;
			else r=m;
		}
		g.inv[i]=l;
	}
}

//†全方位木DP†
//*
typedef struct rerooting_atai{ll a;}atai;
atai dpv[200010];//頂点に対する答え
atai dpe[400010];//辺に対する答え
atai cum[2][400010];//累積和
atai munit={1};//単位元
atai init={-1};
int isinit(atai x){return x.a==init.a;}
atai fold_e(atai x,atai y){
	atai ret;
	ret.a=x.a*y.a%MOD;
	return ret;
}
atai add_v(atai x,int v){return x;}
atai add_e(atai x,int e){x.a++;return x;}

void rrinit(){
	makeinvedge();
	rep(i,0,g.vcnt)dpv[i].a=-1;
	rep(i,0,g.ecnt)dpe[i].a=-1;
}
atai dfs_e(int);
atai dfs_v(int v){
	if(!isinit(dpv[v]))return dpv[v];
	rep(i,g.id[v],g.id[v+1]){
		if(i==g.id[v])cum[0][i]=munit;
		else cum[0][i]=fold_e(cum[0][i-1],dfs_e(i-1));
	}
	for(int i=g.id[v+1]-1;i>=g.id[v];i--){
		if(i==g.id[v+1]-1)cum[1][i]=munit;
		else cum[1][i]=fold_e(cum[1][i+1],dfs_e(i+1));
	}
	atai ret=munit;
	rep(i,g.id[v],g.id[v+1])ret=fold_e(ret,dfs_e(i));
	ret=add_v(ret,v);
	return dpv[v]=ret;
}
atai dfs_e(int e){
	if(!isinit(dpe[e]))return dpe[e];
	atai ret;
	if(!isinit(dpv[g.e[e].g])){
		ret=fold_e(cum[0][g.inv[e]],cum[1][g.inv[e]]);
	}else{
		ret=munit;
		int v=g.e[e].g;
		rep(i,g.id[v],g.id[v+1])if(g.e[i].g!=g.e[e].s)ret=fold_e(ret,dfs_e(i));
	}
	ret=add_v(ret,g.e[e].g);
	ret=add_e(ret,e);
	return dpe[e]=ret;
}
void rerootingDP(int v,int pre){
	if(pre==-1)rrinit();
	dfs_v(v);
	rep(i,g.id[v],g.id[v+1])if(g.e[i].g!=pre)rerootingDP(g.e[i].g,v);
}
//*/

int main(){
	readgraph();
	rerootingDP(0,-1);
	rep(i,0,g.vcnt)printf("%lld\n",dpv[i].a);
}