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 n;
vector<int> G[2009];
int c[2009];
int valasz[2009];
int root;
vector<int> dfs(int x) {
  if (G[x].size() == 0) {
    if (c[x] != 0) {
      cout << "NO" << endl;
      exit(0);
    }
    return vector<int>(1, x);
  }
  vector<int> s;
  for (int sz : G[x]) {
    vector<int> sza = dfs(sz);
    for (int szsz : sza) {
      s.push_back(szsz);
    }
  }
  int mm = s.size();
  if (c[x] > mm) {
    cout << "NO" << endl;
    exit(0);
  }
  s.push_back(x);
  for (int i = mm - 1; i >= c[x]; i--) {
    swap(s[i], s[i + 1]);
  }
  return s;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int pi;
    cin >> pi >> c[i];
    if (pi == 0)
      root = i;
    else
      G[pi].push_back(i);
  }
  vector<int> ans = dfs(root);
  for (int i = 0; i < ans.size(); i++) {
    valasz[ans[i]] = i + 1;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) {
    cout << valasz[i] << " ";
  }
  cout << endl;
  return 0;
}
