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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class B3 {
	static PrintWriter out = new PrintWriter(System.out);

	static int N, root;
	static int[] p, c, sz, curC;
	static long A[];
	static boolean isGood;
	static int child[][];	
	
	public static void main(String[] args) {
		FS in = new FS();
		N = in.nextInt();
		p = new int[N];
		c = new int[N];
		int cnt[] = new int[N];
		
		for(int i = 0; i < N; i++) {
			int par = in.nextInt()-1;
			int cc = in.nextInt();
			p[i] = par;
			if(p[i] == -1) root = i;
			else {
				cnt[p[i]]++;
			}
			c[i] = cc;
		}
		
		child = new int[N][];
		for(int i = 0; i < N; i++) {
			child[i] = new int[cnt[i]];
		}
		
		for(int i = 0; i < N; i++) {
			if(p[i] != -1) {
				child[p[i]][--cnt[p[i]]] = i;
			}
		}
		
		
		isGood = true;
		
		sz = new int[N];
		szdfs(root);

		int x[] = solve(root);
		A = new long[N];
		for(int i = 0; i < N; i++) {
			A[x[i]] = i+1;
		}
		
		
		if(isGood) {
			out.println("YES");
			for(long ll : A) out.print(ll+" ");
			out.println();
		}
		else {
			out.println("NO");
		}
		out.close();
	}
	
	static int[] solve(int node) {
		int res[] = new int[sz[node]];
		
		int idd = 0;
		for(int ii : child[node]) {
			int x[] = solve(ii);
			
			for(int i = 0; i < x.length; i++) {
				// gonna place at idd
				if(idd == c[node]) { res[idd++] = node; }
				res[idd++] = x[i];
			}
		}
		
		if(idd < res.length) {
			if(idd != c[node]) isGood = false;
			res[idd++] = node;
		}
		
		
		return res;
	}
	
	static void szdfs(int n) {
		sz[n] = 1;
		for(int ii : child[n]) {
			szdfs(ii);
			sz[n] += sz[ii];
		}
	}
	
	static class FS{
		BufferedReader br;
		StringTokenizer st;
		public FS() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			while(st == null || !st.hasMoreElements()) {
				try {st = new StringTokenizer(br.readLine());}
				catch(Exception e) { throw null;}
			}
			return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next());}
		double nextDouble() { return Double.parseDouble(next());}
		long nextLong() { return Long.parseLong(next());}
		int[] NIA(int n) {
			int r[] = new int[n];
			for(int i = 0; i < n; i++) r[i] = nextInt();
			return r;
		}
		long[] NLA(int n) {
			long r[] = new long[n];
			for(int i = 0; i < n; i++) r[i] = nextLong();
			return r;
		}
		char[][] grid(int r, int c){
			char res[][] = new char[r][c];
			for(int i = 0; i < r; i++) {
				char l[] = next().toCharArray();
				for(int j = 0; j < c; j++) {
					res[i][j] = l[j];
				}
			}
			return res;
		}
	}
	
}
