Problem:
Score : 800 points

Problem Statement

We have a tree with N vertices. The vertices are numbered 1, 2, ..., N. The i-th (1 ≦ i ≦ N - 1) edge connects the two vertices A_i and B_i.

Takahashi wrote integers into K of the vertices. Specifically, for each 1 ≦ j ≦ K, he wrote the integer P_j into vertex V_j. The remaining vertices are left empty. After that, he got tired and fell asleep.

Then, Aoki appeared. He is trying to surprise Takahashi by writing integers into all empty vertices so that the following condition is satisfied:

Condition: For any two vertices directly connected by an edge, the integers written into these vertices differ by exactly 1.

Determine if it is possible to write integers into all empty vertices so that the condition is satisfied. If the answer is positive, find one specific way to satisfy the condition.

Constraints

1 ≦ N ≦ 10^5

1 ≦ K ≦ N

1 ≦ A_i, B_i ≦ N (1 ≦ i ≦ N - 1)

1 ≦ V_j ≦ N (1 ≦ j ≦ K) (21:18, a mistake in this constraint was corrected)

0 ≦ P_j ≦ 10^5 (1 ≦ j ≦ K)

The given graph is a tree.

All v_j are distinct.

Input

The input is given from Standard Input in the following format:

N
A_1 B_1
A_2 B_2
:
A_{N-1} B_{N-1}
K
V_1 P_1
V_2 P_2
:
V_K P_K

Output

If it is possible to write integers into all empty vertices so that the condition is satisfied, print Yes. Otherwise, print No.

If it is possible to satisfy the condition, print N lines in addition. The v-th (1 ≦ v ≦ N) of these N lines should contain the integer that should be written into vertex v. If there are multiple ways to satisfy the condition, any of those is accepted.

Sample Input 1

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

Sample Output 1

Yes
5
6
6
5
7

The figure below shows the tree when Takahashi fell asleep. For each vertex, the integer written beside it represents the index of the vertex, and the integer written into the vertex is the integer written by Takahashi.

Aoki can, for example, satisfy the condition by writing integers into the remaining vertices as follows:

This corresponds to Sample Output 1. Note that other outputs that satisfy the condition will also be accepted, such as:

Yes
7
6
8
7
7

Sample Input 2

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

Sample Output 2

No

Sample Input 3

4
1 2
2 3
3 4
1
1 0

Sample Output 3

Yes
0
-1
-2
-3

The integers written by Aoki may be negative or exceed 10^6.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.awt.Point;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.PrintWriter;
import java.util.ArrayList;
public class Main
{
    static class Scanner
    {
        BufferedReader br;
        StringTokenizer tk=new StringTokenizer("");
        public Scanner(InputStream is) 
        {
            br=new BufferedReader(new InputStreamReader(is));
        }
        public int nextInt() throws IOException
        {
            if(tk.hasMoreTokens())
                return Integer.parseInt(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextInt();
        }
        public long nextLong() throws IOException
        {
            if(tk.hasMoreTokens())
                return Long.parseLong(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextLong();
        }
        public String next() throws IOException
        {
            if(tk.hasMoreTokens())
                return (tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return next();
        }
        public String nextLine() throws IOException
        {
            tk=new StringTokenizer("");
            return br.readLine();
        }
        public double nextDouble() throws IOException
        {
            if(tk.hasMoreTokens())
                return Double.parseDouble(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextDouble();
        }
        public char nextChar() throws IOException
        {
            if(tk.hasMoreTokens())
                return (tk.nextToken().charAt(0));
            tk=new StringTokenizer(br.readLine());
            return nextChar();
        }
        public int[] nextIntArray(int n) throws IOException
        {
            int a[]=new int[n];
            for(int i=0;i<n;i++)
                a[i]=nextInt();
            return a;
        }
        public long[] nextLongArray(int n) throws IOException
        {
            long a[]=new long[n];
            for(int i=0;i<n;i++)
                a[i]=nextLong();
            return a;
        }
        public int[] nextIntArrayOneBased(int n) throws IOException
        {
            int a[]=new int[n+1];
            for(int i=1;i<=n;i++)
                a[i]=nextInt();
            return a;
        }
        public long[] nextLongArrayOneBased(int n) throws IOException
        {
            long a[]=new long[n+1];
            for(int i=1;i<=n;i++)
                a[i]=nextLong();
            return a;
        }
    
    
    }
    static Exception ex;
    public static void main(String args[]) throws Exception
    {
        Thread r=new Thread(null, new Runnable() {
            public void run() {
                try
                {
                    solve();
                }
                catch(Exception e)
                {
                    ex=e;
                }
            }
        }, "1", 1 << 26);
        r.start();
        r.join();
        if(ex!=null)
            throw ex;
    }
    static ArrayList<Integer> g[];
    static int v[];
    static Point pv[];
    static boolean isValid(Point p){
        if(p.x>p.y)
            return false;
        return true;
    }
    static boolean isInfinite(Point a){
        return a.x==Integer.MIN_VALUE&&a.y==Integer.MAX_VALUE;
    }
    static void merge(Point a, Point b, int chng){
        if(isInfinite(a)&&!isInfinite(b)){
            a.x=b.x-chng;
            a.y=b.y+chng;
            return;
        }else if(isInfinite(b)){
            return;
        }else{
            if(a.x%2!=(b.x-chng)%2){
              a.x=0;a.y=-1;
              return;
            }
            if(((b.x-chng)>a.y)||(a.x>(b.y+chng))){
                a.x=0;
                a.y=-1;
                return;
            }
            a.x=Math.max(a.x,b.x-chng);
            a.y=Math.min(a.y,b.y+chng);
            if(!isValid(a)){
                a.x=0;a.y=-1;
                return;
            }                
        }
    }
    static Point sol(int nd, int p, Point parV){
        //if node already has a value
        if(v[nd]!=0){
            pv[nd]=new Point(v[nd], v[nd]);
            merge(pv[nd], parV,1);
            
        }
        else{
            pv[nd]=new Point(Integer.MIN_VALUE, Integer.MAX_VALUE);
            merge(pv[nd], parV,1);
        }
        if(!isValid(pv[nd]))
                return pv[nd];
        for(int nn:g[nd]){
            if(nn==p)
                continue;
            Point nnv=sol(nn,nd,pv[nd]);
            if(!isValid(nnv))
            {
                pv[nd]=new Point(0,-1);
                break;
            }
            merge(pv[nd],nnv,1);
            if(!isValid(pv[nd]))
                break;
        }
        return pv[nd];
    }
    static void fill(int nd, int p, Point parV){
        //if node doesnot have a value create value
        if(v[nd]==0){
            merge(pv[nd], parV, 1);
            if(isInfinite(pv[nd]))
                v[nd]=0;
            else
                v[nd]=pv[nd].x;
        }
        pv[nd]=new Point(v[nd],v[nd]);
        for(int nn:g[nd]){
            if(nn!=p){
                fill(nn,nd,pv[nd]);
            }
        }
    }
    static void solve() throws IOException
    {
        Scanner in=new Scanner(System.in);
        PrintWriter out=new PrintWriter(System.out);
        int n=in.nextInt();
        g=new ArrayList[n+1];
        for(int i=0;i<=n;i++)
            g[i]=new ArrayList<>();
        for(int i=1;i<n;i++){
            int u=in.nextInt();
            int v=in.nextInt();
            g[u].add(v);
            g[v].add(u);
        }
        v=new int[n+1];
        pv=new Point[n+1];
        for(int i=in.nextInt();i>0;i--){
            v[in.nextInt()]=in.nextInt();
        }
        if(isValid(sol(1,0, new Point(Integer.MIN_VALUE, Integer.MAX_VALUE)))){
            out.println("Yes");
            fill(1,0, new Point(Integer.MIN_VALUE, Integer.MAX_VALUE));
            for(int i=1;i<=n;i++)
                out.println(v[i]);
            
        }
        else{
            out.println("No");
        }
        
        out.close();
        
    }
}