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;
using ll = long long;
vector<int> v[2020], res[2020];
int c[2020];
void h(int i) {
  for (int j : v[i]) {
    h(j);
    if (res[j].size() > res[i].size()) swap(res[i], res[j]);
    for (int x : res[j]) res[i].push_back(x);
  }
  if (c[i] > res[i].size()) {
    cout << "NO\n";
    exit(0);
  }
  res[i].insert(res[i].begin() + c[i], i);
}
int ans[2020];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int r;
  for (int i = 0; i < n; ++i) {
    int p;
    cin >> p >> c[i];
    if (p) {
      p--;
      v[p].push_back(i);
    } else
      r = i;
  }
  h(r);
  int cnt = 1;
  for (int i : res[r]) {
    ans[i] = cnt++;
  }
  cout << "YES\n";
  for (int i = 0; i < n; ++i) cout << ans[i] << " ";
  cout << endl;
}
