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;
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> p) {
  return os << '(' << p.first << ',' << p.second << ')';
}
void init();
void solve();
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T = 1;
  init();
  while (T--) {
    solve();
    cout << '\n';
  }
  return 0;
}
void init() {}
vector<int> trav(vector<vector<int> > m, int r, vector<int> c) {
  vector<int> res(0);
  for (int i = 0; i < m[r].size(); i++) {
    vector<int> tmp = trav(m, m[r][i], c);
    res.insert(res.end(), tmp.begin(), tmp.end());
  }
  if (c[r] > res.size()) {
    cout << "NO";
    exit(0);
  }
  res.insert(next(res.begin(), c[r]), r);
  return res;
}
void solve() {
  int n;
  cin >> n;
  vector<int> p(n + 1), c(n + 1);
  vector<vector<int> > m(n + 1, vector<int>(0));
  for (int i = 1; i <= n; i++) {
    cin >> p[i] >> c[i];
    m[p[i]].push_back(i);
  }
  vector<int> v = trav(m, m[0][0], c);
  ;
  vector<int> val(n + 1, 0);
  for (int i = 0; i < n; i++) {
    val[v[i] - 1] = i + 1;
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) cout << val[i] << " ";
}
