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 long long int mod = 1e9 + 7;
const int inf = 1e9;
const int N = 2000;
vector<int> adj[N + 1];
int c[N + 1], val[N + 1], sub[N + 1];
bool ok = true;
void dfs(int n) {
  for (auto i : adj[n]) {
    dfs(i);
    sub[n] += sub[i];
  }
  sub[n]++;
}
void dfs2(int node, vector<int>& v) {
  if (c[node] > (int)v.size() - 1) {
    ok = false;
    return;
  }
  val[node] = v[c[node]];
  v.erase(v.begin() + c[node]);
  vector<int> temp;
  for (auto i : adj[node]) {
    while ((int)temp.size() != sub[i]) {
      temp.push_back(v.back());
      v.pop_back();
    }
    reverse(temp.begin(), temp.end());
    dfs2(i, temp);
    temp.clear();
  }
}
void solve() {
  int n;
  cin >> n;
  int x, root;
  for (int i = 1; i <= n; i++) {
    cin >> x >> c[i];
    if (x != 0) {
      adj[x].push_back(i);
    } else
      root = i;
  }
  dfs(root);
  vector<int> v;
  for (int i = 1; i <= n; i++) v.push_back(i);
  dfs2(root, v);
  if (!ok)
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (int i = 1; i <= n; i++) {
      cout << val[i] << " ";
    }
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
