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;
constexpr int MAX = 2000 + 5;
vector<int> ch[MAX];
int a[MAX];
int c[MAX];
int r;
int n;
vector<int> solve(int k) {
  vector<int> invalid;
  vector<int> v;
  for (int i : ch[k]) {
    vector<int> w = solve(i);
    if (w.size() == 0) return invalid;
    for (int n : w) {
      v.push_back(n);
    }
  }
  if (c[k] > v.size()) {
    return invalid;
  } else if (c[k] == v.size()) {
    v.push_back(k);
  } else {
    v.insert(v.begin() + c[k], k);
  }
  return v;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p >> c[i];
    if (p == 0) {
      r = i;
    } else {
      ch[p - 1].push_back(i);
    }
  }
  vector<int> v = solve(r);
  if (v.size() == 0) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    for (int i = 0; i < v.size(); i++) {
      a[v[i]] = i + 1;
    }
    for (int i = 0; i < n; i++) {
      cout << a[i] << " ";
    }
    cout << endl;
  }
}
