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 solve(const vector<vector<int>>& edges, const vector<int>& clue,
           vector<int>& iToVal, vector<int>& valToI, int& nextVal, int i) {
  int valBegin = nextVal;
  for (int j : edges[i]) solve(edges, clue, iToVal, valToI, nextVal, j);
  int valEnd = nextVal;
  if (clue[i] > valEnd - valBegin) {
    cout << "NO" << endl;
    exit(0);
  }
  int valI = valBegin + clue[i];
  for (int val = valEnd - 1; val >= valI; --val) {
    int j = valToI[val];
    iToVal[j] = val + 1;
    valToI[val + 1] = j;
  }
  iToVal[i] = valI;
  valToI[valI] = i;
  ++nextVal;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  cin.ignore();
  vector<vector<int>> edges(n);
  vector<int> clue;
  clue.reserve(n);
  int root = 0;
  for (int i = 0; i < n; ++i) {
    int p, c;
    cin >> p >> c;
    cin.ignore();
    if (p == 0)
      root = i;
    else
      edges[p - 1].push_back(i);
    clue.push_back(c);
  }
  vector<int> iToVal(n);
  vector<int> valToI(n + 1);
  int nextVal = 1;
  solve(edges, clue, iToVal, valToI, nextVal, root);
  cout << "YES" << endl;
  for (int val : iToVal) cout << val << ' ';
  cout << endl;
  return 0;
}
