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 ll = long long;
using namespace std;
const ll maxSize = 2007;
struct Node {
  ll value;
  ll ci;
  ll parent;
  vector<ll> childs;
  ll offspring;
};
vector<Node> graph;
bool fillOffspring(ll i, ll ci, const vector<ll>& ids) {
  if (ci >= graph[i].offspring) return false;
  if (graph[i].childs.empty()) {
    graph[i].value = ids.front();
    return true;
  }
  auto it = graph[i].childs.begin();
  auto idIt = ids.begin();
  auto result = true;
  while (it != graph[i].childs.end()) {
    auto child = graph[*it];
    if (ci >= 0 && ci < child.offspring) {
      vector<ll> idCopy(idIt, idIt + child.offspring + 1);
      graph[i].value = *(idCopy.begin() + ci);
      idCopy.erase(idCopy.begin() + ci);
      result = result & fillOffspring(*it, child.ci, idCopy);
      idIt += child.offspring + 1;
    } else {
      vector<ll> idCopy(idIt, idIt + child.offspring);
      result = result & fillOffspring(*it, child.ci, idCopy);
      idIt += child.offspring;
    }
    ci -= child.offspring;
    ++it;
  }
  if (graph[i].value == 0) graph[i].value = ids.back();
  return result;
}
int main() {
  ll n;
  cin >> n;
  graph.resize(n + 1);
  for (ll i = 1; i <= n; ++i) {
    ll pi, ci;
    cin >> pi >> ci;
    graph[i].parent = pi;
    graph[i].ci = ci;
    graph[pi].childs.emplace_back(i);
  }
  for (ll i = 1; i <= n; ++i) {
    vector<bool> visited(n + 1);
    list<ll> queue;
    queue.push_back(i);
    ll offspring = 0;
    while (!queue.empty()) {
      auto next = queue.front();
      queue.pop_front();
      visited[next] = true;
      ++offspring;
      for (const auto& child : graph[next].childs)
        if (!visited[child]) queue.push_back(child);
    }
    graph[i].offspring = offspring;
  }
  vector<ll> ids(n);
  iota(ids.begin(), ids.end(), 1);
  auto root = graph[0].childs.front();
  auto result = fillOffspring(root, graph[root].ci, ids);
  if (result) {
    cout << "YES\n";
    for (ll i = 1; i <= n; ++i) cout << graph[i].value << ' ';
  } else
    cout << "NO";
}
