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;
void numbers_on_tree(int n, vector<int>& p, vector<int>& c) {
  int root = -1;
  function<int(int)> count;
  function<void(int)> construct;
  vector<vector<int> > ch(n + 1);
  vector<int> cnt(n + 1), val(n + 1), r(n);
  for (int i = 1; i <= n; i++) {
    if (p[i] == 0)
      root = i;
    else
      ch[p[i]].push_back(i);
  }
  count = [&](int node) {
    for (int u : ch[node]) cnt[node] += 1 + count(u);
    return cnt[node];
  };
  count(root);
  for (int i = 1; i <= n; i++)
    if (c[i] > cnt[i]) {
      cout << "NO" << endl;
      return;
    }
  iota(r.begin(), r.end(), 1);
  construct = [&](int node) {
    val[node] = r[c[node]];
    r.erase(r.begin() + c[node]);
    for (int u : ch[node]) construct(u);
  };
  construct(root);
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) cout << val[i] << " ";
  cout << endl;
}
int main() {
  int n;
  cin >> n;
  vector<int> p(n + 1), c(n + 1);
  for (int i = 1; i <= n; i++) cin >> p[i] >> c[i];
  numbers_on_tree(n, p, c);
  return 0;
}
