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 main() {
  int n;
  cin >> n;
  vector<int64_t> C(n);
  vector<int> P(n);
  vector<int> deg(n);
  vector<int> proc(n);
  auto children = vector<vector<int>>(n);
  for (int i = 0; i < n; ++i) {
    cin >> P[i] >> C[i];
    --P[i];
    if (P[i] != -1) {
      ++deg[P[i]];
    }
  }
  vector<int64_t> A(n, 1);
  int64_t last = 0;
  stack<int> st;
  for (int i = 0; i < n; ++i) {
    if (deg[i] == 0) st.push(i);
  }
  while (!st.empty()) {
    auto t = st.top();
    st.pop();
    vector<int> children_val;
    children_val.reserve(deg[t] + 1);
    for (auto i : children[t]) {
      children_val.push_back(A[i]);
    }
    sort(children_val.begin(), children_val.end());
    children_val.push_back(++last);
    if (C[t] >= children_val.size()) {
      cout << "NO\n";
      return 0;
    }
    int new_number = children_val[C[t]];
    for (int i = 0; i < n; ++i) {
      if (A[i] >= new_number) {
        ++A[i];
        if (A[i] > last) {
          last = A[i];
        }
      }
    }
    A[t] = new_number;
    if (new_number > last) {
      last = new_number;
    }
    if (P[t] != -1) {
      for (auto i : children[t]) {
        children[P[t]].push_back(i);
      }
      children[P[t]].push_back(t);
      ++proc[P[t]];
      if (proc[P[t]] == deg[P[t]]) {
        st.push(P[t]);
      }
    }
  }
  cout << "YES\n";
  for (auto i : A) {
    cout << i << " ";
  }
  cout << "\n";
}
