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;
int solve(vector<vector<int> > &gr, vector<int> &ord, vector<int> &c, int u) {
  int v;
  int ch = 0;
  int isz = ord.size();
  for (int i = 0; i < gr[u].size(); ++i) {
    v = gr[u][i];
    int tmp = solve(gr, ord, c, v);
    if (tmp == -1) return -1;
    ch += (tmp + 1);
  }
  if (c[u] > ch) return -1;
  auto it = ord.begin();
  if (isz + c[u] >= ord.size())
    it = ord.end();
  else
    it += isz + c[u];
  ord.insert(it, u);
  return ch;
}
int main() {
  int n;
  cin >> n;
  vector<vector<int> > gr(n);
  vector<int> c(n);
  int p;
  int root;
  for (int i = 0; i < n; ++i) {
    cin >> p >> c[i];
    p--;
    if (p != -1) gr[p].push_back(i);
    if (p == -1) root = i;
  }
  vector<int> ord;
  int tmp = solve(gr, ord, c, root);
  if (tmp == -1)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    vector<int> val(n);
    for (int i = 0; i < n; ++i) {
      val[ord[i]] = i + 1;
    }
    for (int i = 0; i < n; ++i) {
      cout << val[i] << " ";
    }
    cout << endl;
  }
  return 0;
}
