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;
bool poss = true;
int n, rt = 0;
vector<int> v, res;
vector<vector<int>> adj;
vector<pair<int, int>> dfs(int second, int p) {
  vector<pair<int, int>> cur;
  for (auto i : adj[second]) {
    if (i != p) {
      vector<pair<int, int>> ch = dfs(i, second);
      for (auto i : ch) {
        cur.push_back(i);
      }
    }
  }
  sort(cur.begin(), cur.end());
  for (int i = 0; i < cur.size(); ++i) {
    if (i < v[second])
      cur[i].first = i;
    else
      cur[i].first = i + 1;
  }
  if (v[second] > cur.size()) poss = false;
  cur.push_back({v[second], second});
  return cur;
}
int main() {
  cin >> n;
  adj.assign(n, vector<int>());
  v.assign(n, 0);
  res.assign(n, 0);
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x >> v[i];
    x--;
    if (x != -1) {
      adj[x].push_back(i);
      adj[i].push_back(x);
    } else
      rt = i;
  }
  vector<pair<int, int>> cur = dfs(rt, rt);
  for (auto i : cur) {
    res[i.second] = i.first + 1;
  }
  if (!poss) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  for (auto i : res) {
    cout << i << " ";
  }
  return 0;
}
