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, p[2020], c[2020];
vector<int> v[2020];
int ans[2020];
vector<pair<int, int>> dfs(int now) {
  if (v[now].size() == 0) {
    if (c[now] != 0) {
      cout << "NO" << endl;
      exit(0);
    }
    ans[now] = 3000 * now;
    vector<pair<int, int>> sol;
    sol.emplace_back(3000 * now, now);
    return sol;
  } else {
    vector<pair<int, int>> sol;
    for (auto to : v[now]) {
      vector<pair<int, int>> tmp = dfs(to);
      for (auto val : tmp) sol.emplace_back(val);
    }
    sort(sol.begin(), sol.end());
    if (c[now] > sol.size()) {
      cout << "NO" << endl;
      exit(0);
    } else {
      int aans;
      if (c[now] == sol.size())
        aans = sol.back().first + 1;
      else
        aans = sol[c[now]].first;
      for (int i = c[now]; i < sol.size(); ++i) {
        sol[i].first++;
        ans[sol[i].second]++;
      }
      sol.emplace_back(aans, now);
      ans[now] = aans;
      return sol;
    }
  }
}
int32_t main() {
  cin.tie(0), cout.sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> p[i] >> c[i];
  int root;
  for (int i = 1; i <= n; ++i) {
    if (p[i] != 0)
      v[p[i]].emplace_back(i);
    else
      root = i;
  }
  dfs(root);
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) {
    cout << ans[i] << " \n"[i == n];
  }
  return 0;
}
