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;
const long long MAXN = 2100;
vector<long long> adj[MAXN];
long long p[MAXN], c[MAXN], j[MAXN];
vector<long long> dfs(long long v) {
  vector<long long> my;
  for (auto u : adj[v]) {
    vector<long long> vv = dfs(u);
    my.insert(my.begin(), vv.begin(), vv.end());
  }
  if (c[v] <= my.size()) {
    my.insert(my.begin() + c[v], v);
  } else {
    cout << "NO";
    exit(0);
  }
  return my;
}
int32_t main() {
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> p[i] >> c[i];
    adj[p[i]].push_back(i);
  }
  c[0] = 0;
  vector<long long> ans = dfs(0);
  for (long long i = 1; i < ans.size(); i++) j[ans[i]] = i;
  cout << "YES\n";
  for (long long i = 1; i < ans.size(); i++) cout << j[i] << ' ';
}
