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;
const int N = 2005;
vector<int> adj[N];
int c[N];
int a[N];
int subsize[N];
void calc_subsize(int u) {
  subsize[u] = 1;
  for (auto child : adj[u]) {
    calc_subsize(child);
    subsize[u] += subsize[child];
  }
}
void dfs(int u, vector<int>& elems) {
  a[u] = elems[c[u]];
  int ptr = 0;
  for (auto child : adj[u]) {
    vector<int> e;
    for (int i = 1; i <= subsize[child]; ++i) {
      if (ptr == c[u]) ptr++;
      e.push_back(elems[ptr++]);
    }
    dfs(child, e);
  }
}
void getAC() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  int root = -1;
  for (int i = 1; i <= n; ++i) {
    int p;
    cin >> p >> c[i];
    if (!p) {
      root = i;
    } else {
      adj[p].push_back(i);
    }
  }
  calc_subsize(root);
  for (int i = 1; i <= n; ++i) {
    if (c[i] >= subsize[i]) {
      cout << "NO" << endl;
      return;
    }
  }
  vector<int> elems;
  for (int i = 1; i <= n; ++i) {
    elems.emplace_back(i);
  }
  dfs(root, elems);
  cout << "YES" << '\n';
  for (int i = 1; i <= n; ++i) cout << a[i] << ' ';
  cout << '\n';
}
int main() {
  getAC();
  return 0;
}
