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;
const long long int inf = 0x3f3f3f3f3f3f3f3fll;
const int maxx = 3000000;
vector<int> adj[2002];
int c[2002], a[2002], ans = 0;
set<pair<int, int> > solve(int v) {
  set<pair<int, int> > papa;
  for (int u : adj[v]) {
    set<pair<int, int> > baccha = solve(u);
    for (pair<int, int> it : baccha) {
      papa.insert({it.first, it.second});
    }
  }
  if (c[v] > papa.size()) {
    ans = -1;
    return set<pair<int, int> >();
  }
  if (papa.size() == 0) {
    papa.insert({1, v});
    return papa;
  }
  int ne = 1;
  set<pair<int, int> > temp;
  int cnt = 0;
  for (auto itr : papa) {
    cnt++;
    if (cnt > c[v])
      temp.insert({itr.first + 2, itr.second});
    else
      temp.insert({itr.first, itr.second});
    if (cnt == c[v]) ne = itr.first + 1;
  }
  temp.insert({ne, v});
  return temp;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int r = -1;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x >> c[i];
    if (x == 0) {
      r = i;
      continue;
    }
    adj[x].push_back(i);
  }
  set<pair<int, int> > st = solve(r);
  if (ans == -1) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  set<pair<int, int> > ans;
  for (auto it : st) ans.insert({it.second, it.first});
  for (auto it : ans) cout << it.second << " ";
  return 0;
}
