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<>();
    		 ans.ensureCapacity(c.length);
    		 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;
    			 }
    		 
    		 ans.add(c[root], root);
    		 return ans;
    	 }
    	
    	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;
    	
    }