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;
long k;
void dfs(long st, list<long> l[], vector<long> v[], long *o, long *c) {
  o[st] = 1;
  for (auto it : l[st]) {
    if (!o[it]) {
      dfs(it, l, v, o, c);
      for (auto i : v[it]) v[st].push_back(i);
    }
  }
  if (c[st] > v[st].size()) {
    k = 1;
    return;
  }
  v[st].insert(v[st].begin() + c[st], st);
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long n;
  cin >> n;
  long *c = new long[n];
  long *o = (long *)calloc(n, sizeof(long));
  list<long> l[n];
  long root;
  for (long i = 0; i < n; i++) {
    long p;
    cin >> p >> c[i];
    if (p > 0) {
      p--;
      l[p].push_back(i);
      l[i].push_back(p);
    } else
      root = i;
  }
  vector<long> v[n];
  dfs(root, l, v, o, c);
  if (k == 1)
    cout << "NO";
  else {
    cout << "YES\n";
    long *ans = new long[n];
    for (long i = 0; i < n; i++) {
      ans[v[root][i]] = i + 1;
    }
    for (long i = 0; i < n; i++) {
      cout << ans[i] << " ";
    }
    delete[] ans;
  }
  delete[] c;
  delete[] o;
  return 0;
}
