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
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF1287D extends PrintWriter {
	CF1287D() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1287D o = new CF1287D(); o.main(); o.flush();
	}

	int n;
	int[] eo; int[][] ej;
	int[] cc, sz, aa;
	boolean[] used;
	void append(int i, int j) {
		int o = eo[i]++;
		if (o >= 2 && (o & o - 1) == 0)
			ej[i] = Arrays.copyOf(ej[i], o << 1);
		ej[i][o] = j;
	}
	int query(int c) {
		for (int a = 1; a <= n; a++)
			if (!used[a] && c-- == 0)
				return a;
		return 0;
	}
	boolean dfs(int i) {
		int a = query(cc[i]);
		if (a == 0)
			return false;
		used[a] = true;
		aa[i] = a;
		sz[i] = 1;
		for (int o = eo[i]; o-- > 0; ) {
			int j = ej[i][o];
			if (!dfs(j))
				return false;
			sz[i] += sz[j];
		}
		return cc[i] < sz[i];
	}
	void main() {
		n = sc.nextInt();
		eo = new int[n]; ej = new int[n][2];
		cc = new int[n]; sz = new int[n]; aa = new int[n];
		used = new boolean[n + 1];
		int r = -1;
		for (int j = 0; j < n; j++) {
			int i = sc.nextInt() - 1;
			int c = sc.nextInt();
			if (i == -1)
				r = j;
			else
				append(i, j);
			cc[j] = c;
		}
		if (dfs(r)) {
			println("YES");
			for (int i = 0; i < n; i++)
				print(aa[i] + " ");
			println();
		} else
			println("NO");
	}
}
