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 main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<vector<int> > adj(n);
  vector<int> c(n);
  int root;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p >> c[i];
    --p;
    if (p == -1) {
      root = i;
    } else {
      adj[p].push_back(i);
    }
  }
  vector<int> a(n);
  int val = 1;
  vector<pair<int, int> > vec;
  function<void(int)> dfs = [&](int u) {
    vec.emplace_back(a[u], u);
    for (auto v : adj[u]) {
      dfs(v);
    }
  };
  bool can = true;
  function<void(int)> find_ans = [&](int u) {
    for (auto v : adj[u]) {
      find_ans(v);
    }
    if (adj[u].size() == 0) {
      if (c[u] != 0)
        can = false;
      else
        a[u] = val++;
    } else {
      vec.clear();
      for (auto v : adj[u]) {
        dfs(v);
      }
      sort(vec.begin(), vec.end());
      if (c[u] > vec.size()) {
        can = false;
      } else {
        if (c[u] == vec.size()) {
          a[u] = vec.back().first + 1;
        } else {
          a[u] = vec[c[u]].first + 1;
          for (int i = c[u]; i < vec.size(); i++) {
            a[vec[i].second] += 2;
          }
        }
      }
    }
  };
  find_ans(root);
  if (can) {
    cout << "YES" << endl;
    for (auto x : a) cout << x << " ";
    cout << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}
