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;
int n;
map<int, vector<int> > m;
vector<int> s;
vector<int> c;
vector<int> ans;
void dfs(int i) {
  if (s.size() <= c[i]) {
    return;
  }
  ans[i] = s[c[i]];
  s.erase(s.begin() + c[i]);
  for (auto it : m[i]) {
    dfs(it);
  }
  if (!s.empty()) {
    if (*s.begin() < ans[i]) {
      ans[i] = 0;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(10);
  int Test = 1;
  while (Test--) {
    cin >> n;
    int root;
    ans.resize(n, 0);
    for (int i = 0; i < n; i++) {
      s.push_back(i + 1);
      int x, y;
      cin >> x >> y;
      c.push_back(y);
      if (x == 0)
        root = i;
      else
        m[x - 1].push_back(i);
    }
    dfs(root);
    for (int i = 0; i < n; i++) {
      if (ans[i] == 0) {
        cout << "NO" << endl;
        return 0;
      }
    }
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) cout << ans[i] << " ";
  }
}
