Evlampiy was gifted a rooted tree. The vertices of the tree are numbered from 1 to n. Each of its vertices also has an integer a_i written on it. For each vertex i, Evlampiy calculated c_i — the number of vertices j in the subtree of vertex i, such that a_j < a_i. 

<image>Illustration for the second example, the first integer is a_i and the integer in parentheses is c_i

After the new year, Evlampiy could not remember what his gift was! He remembers the tree and the values of c_i, but he completely forgot which integers a_i were written on the vertices.

Help him to restore initial integers!

Input

The first line contains an integer n (1 ≤ n ≤ 2000) — the number of vertices in the tree.

The next n lines contain descriptions of vertices: the i-th line contains two integers p_i and c_i (0 ≤ p_i ≤ n; 0 ≤ c_i ≤ n-1), where p_i is the parent of vertex i or 0 if vertex i is root, and c_i is the number of vertices j in the subtree of vertex i, such that a_j < a_i.

It is guaranteed that the values of p_i describe a rooted tree with n vertices.

Output

If a solution exists, in the first line print "YES", and in the second line output n integers a_i (1 ≤ a_i ≤ {10}^{9}). If there are several solutions, output any of them. One can prove that if there is a solution, then there is also a solution in which all a_i are between 1 and 10^9.

If there are no solutions, print "NO".

Examples

Input


3
2 0
0 2
2 0


Output


YES
1 2 1 

Input


5
0 1
1 3
2 1
3 0
2 0


Output


YES
2 3 2 1 2
import java.util.*;
public class Main {
	static int pre[]=new int [5000];
	static int c[]=new int [5000];
	static Vector <Integer>vec[]=new Vector[5000];
	static int son[]=new int [5000];
	static int vis[]=new int[5000];
	static int n;
	static int ans[]=new int [5000];
	 static void dfs(int root)
	{
		if(vec[root].size()==0)
		{
			son[root]=0;
			return ;
		}
		else
		for(int i=0;i<vec[root].size();i++)
		{
			if(pre[root]!=vec[root].get(i))
			dfs(vec[root].get(i));
			else continue;
			son[root]+=(1+son[vec[root].get(i)]);
		}
	}
	static void dfs2(int root)
	{
		int cnt=0;
		for(int i=1;i<=n;i++)
		{
			if(vis[i]==0)cnt++;
			if(c[root]+1==cnt)
			{
				ans[root]=i;vis[i]=1;
				break;
			}
		}
		for(int i=0;i<vec[root].size();i++)
		{
			if(vec[root].get(i)==pre[root])continue;
			dfs2(vec[root].get(i));
		}
	}
 public static void main(String args[])
 {
	// System.out.println("nmsl");
	 Scanner input=new Scanner(System.in);
	n=input.nextInt();
	 for(int i=0;i<=n;i++)
		 vec[i]=new Vector<Integer>();
	 int root=0;
	 for(int i=1;i<=n;i++)
	 {
		 int p;
		 p=input.nextInt();c[i]=input.nextInt();
		 if(p==0)root=i;
		 pre[i]=p;vec[p].add(i);vec[i].add(p);
	 }
	 dfs(root);boolean f=true;
	 for(int i=1;i<=n;i++)
	 {
		 if(c[i]>son[i])
		 {
			 f=false;break;
		 }
	 }
	 if(!f)System.out.println("NO");
	 else
	 {
		 System.out.println("YES");
		 dfs2(root);
		 for(int i=1;i<=n;i++)
			 System.out.print(ans[i]+" ");
	 }
 }
}
 	  		  		 			 		  					 				