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
#include <bits/stdc++.h>
using namespace std;
struct Node {
  int c;
  vector<int> child;
};
Node Tree[2000];
bool calculate_a(int root, vector<int>& sorted_node) {
  vector<int>& children = Tree[root].child;
  for (int i = 0; i < children.size(); i++) {
    int child = children[i];
    vector<int> child_sorted_node;
    if (!calculate_a(child, child_sorted_node)) return false;
    sorted_node.insert(sorted_node.end(), child_sorted_node.begin(),
                       child_sorted_node.end());
  }
  int insert_poi = Tree[root].c;
  if (insert_poi > sorted_node.size()) {
    return false;
  }
  sorted_node.insert(sorted_node.begin() + insert_poi, root);
  ;
  return true;
}
int main() {
  istream& data_stream = cin;
  int node_num;
  data_stream >> node_num;
  for (int i = 1; i < node_num + 1; i++) {
    int parent;
    int c;
    data_stream >> parent >> c;
    Tree[parent].child.push_back(i);
    Tree[i].c = c;
  }
  vector<int> sorted_node;
  if (!calculate_a((Tree[0].child)[0], sorted_node)) {
    printf("NO");
    return 0;
  }
  printf("YES\n");
  int* table = new int[node_num + 1];
  for (int i = 0; i < sorted_node.size(); i++) {
    table[sorted_node[i]] = i + 1;
  }
  for (int i = 1; i < node_num + 1; i++) {
    printf("%d ", table[i]);
  }
  return 0;
}
