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;
long long mod(long long a, long long b) {
  return a >= 0 ? a % b : (b + a % b) % b;
}
long long dceil(long long a, long long b) { return (a + b - 1) / b; }
vector<int> adjs[2000];
bool visited[2000];
int cs[2000];
int res[2000];
bool dfsRec(int cv, vector<int> &order) {
  visited[cv] = true;
  bool pushed = false;
  for (int i = 0; i < adjs[cv].size(); i++) {
    int av = adjs[cv][i];
    if (!visited[av]) {
      vector<int> subtree;
      if (!dfsRec(av, subtree)) return false;
      for (auto it = subtree.begin(); it != subtree.end();) {
        if (order.size() == cs[cv]) {
          order.push_back(cv);
          pushed = true;
        } else {
          order.push_back(*it);
          it++;
        }
      }
    }
  }
  if (order.size() == cs[cv])
    order.push_back(cv);
  else if (!pushed)
    return false;
  return true;
}
bool dfs(int r, int v) {
  for (int i = 0; i < v; i++) {
    visited[i] = false;
  }
  vector<int> tree;
  if (dfsRec(r, tree)) {
    int i = 1;
    for (auto it = tree.begin(); it != tree.end(); it++) {
      res[*it] = i;
      i++;
    }
    return true;
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int root;
  for (int i = 0; i < n; i++) {
    int p, c;
    cin >> p >> cs[i];
    p--;
    if (p != -1)
      adjs[p].push_back(i);
    else
      root = i;
  }
  if (dfs(root, n)) {
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      cout << res[i] << " ";
    }
    cout << "\n";
  } else
    cout << "NO\n";
  return 0;
}
