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;
int root, n, val[2010], res[2010];
int f;
vector<int> edge[2010];
queue<int> dfs(int now) {
  queue<int> temp;
  for (auto to : edge[now]) {
    queue<int> buf = dfs(to);
    while (!buf.empty()) {
      temp.push(buf.front());
      buf.pop();
    }
  }
  int sz = temp.size();
  if (val[now] > sz) f = 1;
  if (val[now] == 0) temp.push(now);
  for (int i = 1; i <= sz; i++) {
    temp.push(temp.front());
    temp.pop();
    if (val[now] == i) temp.push(now);
  }
  return temp;
}
int main() {
  cin >> n;
  for (int i = 1, fa; i <= n; i++) {
    cin >> fa >> val[i];
    if (fa != 0)
      edge[fa].push_back(i);
    else
      root = i;
  }
  queue<int> que = dfs(root);
  int cnt = 0;
  while (!que.empty()) {
    res[que.front()] = ++cnt;
    que.pop();
  }
  if (f) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << res[i] << " ";
  return 0;
}
