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.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String[] line;
        line = reader.readLine().split(" ");
        int n = Integer.parseInt(line[0]);
        V[] tree = new V[n];
        for (int i = 0; i < n; i++) {
            line = reader.readLine().split(" ");
            tree[i] = new V(i, Integer.parseInt(line[0]) - 1, Integer.parseInt(line[1]));
        }

        V root = null;
        for (int i = 0; i < n; i++) {
            if (tree[i].p == -1) {
                root = tree[i];
            } else {
                tree[tree[i].p].addTo(tree[i]);
            }
        }

        ArrayList<V> order = getOrder(root);
        if (order == null) {
            System.out.println("NO");
        } else {
            System.out.println("YES");
            int[] res = new int[n];
            for (int i = 0; i < n; i++) {
                res[order.get(i).index] = i + 1;
            }
            StringBuilder s = new StringBuilder();
            for (int i = 0; i < n; i++) {
                s.append(res[i] + " ");
            }
            System.out.println(s.toString());
        }
    }

    private static ArrayList<V> getOrder(V v) {
        if (v == null) {
            return new ArrayList<>();
        }
        ArrayList<V> order = new ArrayList<>();
        int c = 0;
        for (V to : v.to) {
            ArrayList<V> orderTo = getOrder(to);
            if (orderTo == null) {
                return null;
            }
            for (V orTo: orderTo) {
                if (c == v.c) {
                    order.add(v);
                }
                order.add(orTo);
                c++;
            }
        }
        if (c == v.c) {
            order.add(v);
        } else if (c < v.c) {
            return null;
        }
        return order;
    }

    public static class V {
        public V(int index, int p, int c) {
            this.index = index;
            this.c = c;
            this.p = p;
        }

        public void addTo(V v) {
            to.add(v);
        }
        int index;
        int c;
        int p;
        ArrayList<V> to = new ArrayList<>();
    }
}