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;
constexpr long long int MOD = 1000000007;
constexpr double EPS = 1e-8;
long long int N, M, K, H, W, L, R;
bool ed = false;
void Search(vector<vector<int>>& edge, vector<vector<int>>& p, vector<int>& num,
            int node) {
  if (ed) return;
  for (auto i : edge[node]) {
    Search(edge, p, num, i);
    if (ed) return;
    for (auto j : p[i]) p[node].push_back(j);
  }
  if (p[node].size() < num[node]) {
    ed = true;
    return;
  }
  p[node].insert((p[node].begin() + num[node]), node);
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  vector<int> parent(N);
  vector<vector<int>> edge(N);
  vector<vector<int>> p(N);
  vector<int> num(N);
  int root = 0;
  for (int i = 0; i < N; i++) {
    cin >> L >> R;
    L--;
    parent[i] = L;
    if (L >= 0)
      edge[L].push_back(i);
    else
      root = i;
    num[i] = R;
  }
  Search(edge, p, num, root);
  if (ed) {
    cout << "NO\n";
    return 0;
  }
  vector<int> ans(N);
  for (int i = 0; i < N; i++) {
    ans[p[root][i]] = i;
  }
  cout << "YES\n";
  for (auto i : ans) cout << i + 1 << " ";
  cout << endl;
}
