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;
vector<vector<int>> adja_list;
unordered_map<int, int> c_map;
unordered_map<int, int> ans;
int counter = 1;
int fish(int cur) {
  int seed = counter;
  for (auto &x : adja_list[cur]) {
    int sub_seed = fish(x);
    if (sub_seed == -1) return -1;
    seed = min(seed, sub_seed);
  }
  if (c_map[cur] > counter - seed) {
    return -1;
  }
  for (int i = counter; i >= c_map[cur] + seed; i--) {
    ans[i + 1] = ans[i];
  }
  ans[c_map[cur] + seed] = cur;
  counter++;
  return seed;
}
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  adja_list.assign(n, vector<int>());
  int root = -1;
  for (int i = 0; i < n; i++) {
    int p, c;
    cin >> p >> c;
    p--;
    c_map[i] = c;
    if (p == -1) {
      root = i;
      continue;
    }
    adja_list[p].push_back(i);
  }
  int rez = fish(root);
  if (rez == -1) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    unordered_map<int, int> true_ans;
    for (auto &x : ans) {
      true_ans[x.second] = x.first;
    }
    for (int i = 0; i < n; i++) {
      cout << true_ans[i] << ' ';
    }
  }
  return 0;
}
