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;
void dfs2(vector<vector<pair<int, int> > > &v, int node, vector<int> &vi,
          vector<pair<int, int> > &child) {
  int n = v[node].size();
  for (int i = 0; i < n; i++) dfs2(v, v[node][i].first, vi, child);
  child.push_back(make_pair(vi[node], node));
}
bool dfs1(vector<vector<pair<int, int> > > &v, vector<int> &vi,
          pair<int, int> &node, int &x) {
  int n = v[node.first].size();
  if (n == 0 && node.second) return false;
  if (n == 0) {
    vi[node.first] = x++;
    return true;
  }
  vector<pair<int, int> > child;
  for (int i = 0; i < n; i++) {
    if (!dfs1(v, vi, v[node.first][i], x)) return false;
  }
  x++;
  for (int i = 0; i < n; i++) {
    dfs2(v, v[node.first][i].first, vi, child);
  }
  if (child.size() < node.second) return false;
  sort(child.begin(), child.end());
  if (node.second == 0)
    vi[node.first] = child[0].first;
  else
    vi[node.first] = child[node.second - 1].first + 1;
  for (int i = node.second; i < child.size(); i++) {
    ++vi[child[i].second];
  }
  return true;
}
int main() {
  int n;
  cin >> n;
  pair<int, int> p;
  vector<vector<pair<int, int> > > v(n);
  for (int i = 0; i < n; i++) {
    int u, c;
    cin >> u >> c;
    if (u == 0)
      p = make_pair(i, c);
    else
      v[u - 1].push_back(make_pair(i, c));
  }
  vector<int> vi(n);
  int x = 1;
  if (!dfs1(v, vi, p, x)) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) cout << vi[i] << " ";
  cout << endl;
  return 0;
}
