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;
vector<int> tree[2010];
int cs[2010], ps[2010];
int numDesc(int node) {
  int descendants = 0;
  for (int x : tree[node]) {
    descendants += 1 + numDesc(x);
  }
  if (descendants < cs[node]) return 10000;
  return descendants;
}
void solve(int node, vector<int>& order) {
  ps[node] = order[cs[node]];
  order.erase(order.begin() + cs[node]);
  for (int x : tree[node]) solve(x, order);
}
int main() {
  int n;
  cin >> n;
  int root;
  for (int i = 1; i <= n; i++) {
    int p, c;
    cin >> p >> c;
    cs[i] = c;
    if (p == 0) {
      root = i;
      continue;
    }
    tree[p].push_back(i);
  }
  if (numDesc(root) > 5000)
    cout << "NO";
  else {
    vector<int> order(n);
    order[0] = 1;
    for (int i = 1; i < n; i++) order[i] = order[i - 1] + 1;
    solve(root, order);
    cout << "YES\n";
    for (int i = 1; i < n; i++) cout << ps[i] << " ";
    cout << ps[n];
  }
  return 0;
}
