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;
vector<long long> v[2005];
queue<long long> q;
long long t[2005], c[2005], val[2005];
void solve(int root) {
  int act = 1, nod;
  while (act <= n) {
    while (!q.empty()) q.pop();
    q.push(root);
    while (1) {
      if (q.empty()) {
        cout << "NO" << '\n';
        exit(0);
      }
      nod = q.front();
      q.pop();
      if (c[nod] == 0 && val[nod] == -1) {
        val[nod] = act;
        act++;
        for (int x = t[nod]; x != 0; x = t[x]) c[x]--;
        break;
      }
      for (auto vec : v[nod]) q.push(vec);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> t[i] >> c[i];
    v[t[i]].push_back(i);
    val[i] = -1;
  }
  val[0] = -1;
  c[0] = 1000000000000;
  solve(0);
  cout << "YES" << '\n';
  for (int i = 1; i <= n; i++) cout << val[i] << ' ';
  return 0;
}
