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
// package Quarantine;

import javafx.util.Pair;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class NumbersOnATree {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(br.readLine());
        int par[]=new int[n+1];
        int c[]=new int[n+1];
        StringTokenizer st;
        int root=0;
        ArrayList<Integer> tree[]=new ArrayList[n+1];
        for(int i=1;i<=n;i++){
            tree[i]=new ArrayList<>();
        }
        for(int i=1;i<=n;i++){
            st=new StringTokenizer(br.readLine());
            par[i]=Integer.parseInt(st.nextToken());
            c[i]=Integer.parseInt(st.nextToken());
            if(par[i]==0){
                root=i;
            }
            else{
                tree[par[i]].add(i);
                tree[i].add(par[i]);
            }
        }
        int map[]=new int[n+1];
        dfs2(tree,root,new boolean[n+1],c,map);
        StringBuilder print=new StringBuilder();
        print.append("YES\n");
        for(int i=1;i<=n;i++){
            print.append(map[i]+" ");
        }
        System.out.print(print.toString());
    }
    public static ArrayList<Integer> dfs2(ArrayList<Integer> tree[],int curr,
                            boolean visited[],int c[],int map[]){
        visited[curr]=true;
        int ahead=c[curr];
        ArrayList<Integer> children=new ArrayList<>();
        int p=1;
        for(int j:tree[curr]){
            if(!visited[j]){
                ArrayList<Integer> temp=dfs2(tree,j,visited,c,map);
                for(int k=0;k<temp.size();k++){
                    children.add(temp.get(k));
                    map[temp.get(k)]=p;
                    p++;
                }
            }
        }
        if(children.size()<ahead){
            System.out.println("NO");
            System.exit(0);
        }
        ArrayList<Integer> ans=new ArrayList<>();
        for(int i=0;i<ahead;i++){
            ans.add(children.get(i));
        }
        ans.add(curr);
        map[curr]=ahead+1;
        for(int i=ahead;i<children.size();i++){
            ans.add(children.get(i));
            map[children.get(i)]++;
        }
        return ans;
    }
}
