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 Fast_IO() {
  ios::sync_with_stdio(0);
  cin.tie(0);
}
long long n, root, poss;
long long parent[2020], want[2020], val[2020];
vector<vector<long long> > adj;
long long assigner[2020];
long long subsize[2020];
void dfs(long long u) {
  subsize[u] = 1;
  for (long long i = (long long)0; i <= (long long)int(adj[u].size()) - 1; i++)
    if (adj[u][i] != parent[u]) dfs(adj[u][i]);
  for (long long i = (long long)0; i <= (long long)int(adj[u].size()) - 1; i++)
    if (adj[u][i] != parent[u]) subsize[u] += subsize[adj[u][i]];
}
void dfs2(long long u) {
  if (want[u] > subsize[u] - 1) {
    poss = 0;
    return;
  }
  val[u] = assigner[want[u]];
  for (long long i = (long long)want[u]; i <= (long long)2018; i++)
    assigner[i] = assigner[i + 1];
  for (long long i = (long long)0; i <= (long long)int(adj[u].size()) - 1; i++)
    if (adj[u][i] != parent[u]) dfs2(adj[u][i]);
}
int main() {
  Fast_IO();
  poss = 1;
  for (long long i = (long long)0; i <= (long long)2019; i++)
    assigner[i] = i + 1;
  cin >> n;
  for (long long i = (long long)0; i <= (long long)n - 1; i++)
    cin >> parent[i] >> want[i];
  for (long long i = (long long)0; i <= (long long)n - 1; i++) parent[i]--;
  adj.resize(n);
  for (long long i = (long long)0; i <= (long long)n - 1; i++) {
    if (parent[i] == -1)
      root = i;
    else {
      adj[i].push_back(parent[i]);
      adj[parent[i]].push_back(i);
    }
  }
  dfs(root);
  dfs2(root);
  if (!poss)
    cout << "NO" << '\n';
  else {
    cout << "YES" << '\n';
    for (long long i = (long long)0; i <= (long long)n - 1; i++)
      cout << val[i] << " ";
  }
  return 0;
}
