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, c[2003], f[2003];
vector<int> graph[2003];
vector<int> dfs(int node) {
  vector<int> res;
  for (auto &nxt : graph[node]) {
    vector<int> sv = dfs(nxt);
    for (auto &c : sv) {
      res.push_back(c);
    }
  }
  res.push_back(node);
  int idx = (int)res.size() - 1;
  if (c[node] > idx) {
    cout << "NO\n";
    exit(0);
  }
  while (idx > c[node]) {
    swap(res[idx - 1], res[idx]);
    idx--;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  int root = -1;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p >> c[i + 1];
    graph[p].push_back(i + 1);
    if (p == 0) root = i + 1;
  }
  vector<int> ans = dfs(root);
  for (int i = 0; i < n; i++) {
    f[ans[i]] = i + 1;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) {
    if (i > 1) cout << ' ';
    cout << f[i];
  }
  cout << '\n';
  return 0;
}
