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 test1 {

	static HashSet<Integer>[]gr;
	static int []c;
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int t =1;
		StringBuilder st = new StringBuilder();

		while (t-- > 0) {
		
			int n=sc.nextInt();
			 c=new int[n+1];
			int root=0;
			gr=new HashSet[n+1];
			for(int i=0;i<=n;i++) {
				gr[i]=new HashSet<>();
			}
			for(int i=1;i<=n;i++) {
				int in=sc.nextInt();
				c[i]=sc.nextInt();
				if(in==0) {
					root=i;
				}
				else {
					gr[in].add(i);
				}
			}
			
			ArrayList<Integer> res=dfs(root);
			int a[]=new int [n+1];
			if(!f)
				System.out.println("NO");
			else {
				System.out.println("YES");
				for(int i=0;i<n;i++) {
					a[res.get(i)]=i+1;
					
				}
				for(int i=1;i<=n;i++) {
					System.out.print(a[i]+" ");
				}
			}
			
			
		}
		System.out.println(st);

		
	}
	static boolean f=true;
	 static ArrayList<Integer> dfs(int root){
		 ArrayList<Integer> ans=new ArrayList<>();
		 ArrayList<Integer> al=new ArrayList<>();
		 
		 for(Integer ch:gr[root]) {
			 ArrayList<Integer> res=dfs(ch);
			 ans.addAll(res);
		 }
		 if(f==false)return ans;
			 if(c[root]>ans.size()) {
				 f=false;
				 return ans;
			 }
		 int i=0;
		 for( i=0;i<c[root];i++) {
			 al.add(ans.get(i));
		 }
		 al.add(root);
		 while(i<ans.size()) {
			 al.add(ans.get(i));
			 i++;
		 }
		 return al;
	 }
	
	static boolean prime[] = new boolean[1000001]; 
	static void sieve() 
    { 
       
        
        for(int i=0;i<1000001;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <=1000000; p++) 
        { 
            // If prime[p] is not changed, then it is a prime 
            if(prime[p] == true) 
            { 
                // Update all multiples of p 
                for(int i = p*p; i <= 1000000; i += p) 
                    prime[i] = false; 
            } 
        } 
          
        // Print all prime numbers 
        for(int i = 2; i <= 1000000; i++) 
        { 
            //if(prime[i] == true) 
               
        } 
    } 
	static int cnt=0;
	
}
