Problem: Alice and Bob are playing a game. They have a tree consisting of n vertices. Initially, Bob has k chips, the i-th chip is located in the vertex a_i (all these vertices are unique). Before the game starts, Alice will place a chip into one of the vertices of the tree.

The game consists of turns. Each turn, the following events happen (sequentially, exactly in the following order):

  1. Alice either moves her chip to an adjacent vertex or doesn't move it; 
  2. for each Bob's chip, he either moves it to an adjacent vertex or doesn't move it. Note that this choice is done independently for each chip. 



The game ends when Alice's chip shares the same vertex with one (or multiple) of Bob's chips. Note that Bob's chips may share the same vertex, even though they are in different vertices at the beginning of the game.

Alice wants to maximize the number of turns, Bob wants to minimize it. If the game ends in the middle of some turn (Alice moves her chip to a vertex that contains one or multiple Bob's chips), this turn is counted.

For each vertex, calculate the number of turns the game will last if Alice places her chip in that vertex.

Input

The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of vertices in the tree.

Then n - 1 lines follow, each line contains two integers u_i, v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i) that denote the endpoints of an edge. These edges form a tree.

The next line contains one integer k (1 ≤ k ≤ n - 1) — the number of Bob's chips.

The last line contains k integers a_1, a_2, ..., a_k (1 ≤ a_i ≤ n; a_i ≠ a_j if i ≠ j) — the vertices where the Bob's chips are initially placed.

Output

Print n integers. The i-th of them should be equal to the number of turns the game will last if Alice initially places her chip in the vertex i. If one of Bob's chips is already placed in vertex i, then the answer for vertex i is 0.

Examples

Input


5
2 4
3 1
3 4
3 5
2
4 5


Output


2 1 2 0 0


Input


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


Output


3 0 0 3 1 2 3 0


Input


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


Output


0 2 2 2 2 0 2 2 0 0

Solution (in CPP):

Solution:
#include <algorithm>  
#include <iostream>  
#include <sstream>  
#include <string>  
#include <vector>  
#include <queue>  
#include <set>  
#include <map>  
#include <cstdio>  
#include <cstdlib>  
#include <cctype>  
#include <cmath>  
#include <cstring>
#include <list>  
#include <cassert>
#include <climits>
#include <bitset>
#include <chrono>
#include <random>
using namespace std;

#define PB push_back  
#define MP make_pair  
#define SZ(v) ((int)(v).size())  
#define FOR(i,a,b) for(int i=(a);i<(b);++i)  
#define REP(i,n) FOR(i,0,n)  
#define FORE(i,a,b) for(int i=(a);i<=(b);++i)  
#define REPE(i,n) FORE(i,0,n)  
#define FORSZ(i,a,v) FOR(i,a,SZ(v))  
#define REPSZ(i,v) REP(i,SZ(v))  
std::mt19937 rnd((int)std::chrono::steady_clock::now().time_since_epoch().count());
typedef long long ll;
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }

const int MAXN = 200000;

int n;
vector<int> adj[MAXN];
int nbob;
int bob[MAXN];
int ans[MAXN];

int dbob[MAXN];

void calcdbob() {
	queue<int> q;
	REP(i, n) dbob[i] = INT_MAX;
	REP(i, nbob) { int at = bob[i]; dbob[at] = 0; q.push(at); }
	while (!q.empty()) {
		int at = q.front(); q.pop();
		REPSZ(i, adj[at]) { int to = adj[at][i]; if (dbob[to] == INT_MAX) dbob[to] = dbob[at] + 1, q.push(to); }
	}
}

bool alive[MAXN];
int sz[MAXN];

void dfssz(int at, int par) {
	sz[at] = 1;
	REPSZ(i, adj[at]) {
		int to = adj[at][i];
		if (to == par || !alive[to]) continue;
		dfssz(to, at);
		sz[at] += sz[to];
	}
}
int dfscentroid(int at, int par, int rootsz) {
	REPSZ(i, adj[at]) {
		int to = adj[at][i];
		if (to == par || !alive[to]) continue;
		if (2 * sz[to] > rootsz) return dfscentroid(to, at, rootsz);
	}
	return at;
}
int centroid(int at) {
	dfssz(at,-1);
	return dfscentroid(at, -1, sz[at]);
}

int lst[MAXN], nlst;
int par[MAXN], dep[MAXN];
int mxdep;
int best[MAXN + 1];

void dfsinit(int at) {
	lst[nlst++] = at;
	dep[at] = par[at] == -1 ? 0 : dep[par[at]] + 1;
	REPSZ(i, adj[at]) {
		int to = adj[at][i];
		if (to == par[at] || !alive[to]) continue;
		par[to] = at;
		dfsinit(to);
	}
}

void processcentroid(int cent,int lvl) {
	//printf("processing %d at lvl %d\n", cent, lvl);
	nlst = 0; par[cent] = -1; dfsinit(cent);
	mxdep = 0; REP(i, nlst) mxdep = max(mxdep, dep[lst[i]]);
	REPE(d, mxdep) best[d] = INT_MIN;
	REP(i, nlst) {
		int at = lst[i];
		int d = min(mxdep, dbob[at] - dep[at] - 1);
		if (d < 0) continue;
		//printf("\tto %d: mxdep=%d -> %d\n", at, d, dbob[at]);
		best[d] = max(best[d], dbob[at]);
	}
	for (int d = mxdep - 1; d >= 0; --d) best[d] = max(best[d], best[d + 1]);
	REP(i, nlst) {
		int at = lst[i];
		ans[at] = max(ans[at], best[dep[at]]);
	}
}

void decompose(int at,int lvl=0) {
	at = centroid(at);
	processcentroid(at, lvl);
	alive[at] = false;
	REPSZ(i, adj[at]) {
		int to = adj[at][i];
		if (!alive[to]) continue;
		decompose(to, lvl + 1);
	}
}

void solve() {
	calcdbob();
	REP(i, n) ans[i] = 0;
	REP(i, n) alive[i] = true;
	decompose(0);
}

void run() {
	scanf("%d", &n);
	REP(i, n - 1) { int a, b; scanf("%d%d", &a, &b); --a, --b; adj[a].PB(b), adj[b].PB(a); }
	scanf("%d", &nbob);
	REP(i, nbob) scanf("%d", &bob[i]), --bob[i];
	solve();
	REP(i, n) { if (i != 0) printf(" "); printf("%d", ans[i]); } puts("");
}

int main() {
	run();
	return 0;
}