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 MXN = 2001;
int n;
int c[MXN];
vector<int> adj[MXN];
bool vis[MXN];
int start = 1;
bool imposs = false;
list<pair<int, int> > dfs(int cur) {
  list<pair<int, int> > now;
  vis[cur] = true;
  for (int x : adj[cur]) {
    if (vis[x]) continue;
    list<pair<int, int> > tmp = dfs(x);
    for (auto it = tmp.begin(); it != tmp.end(); it++)
      now.insert(now.end(), *it);
    if (imposs) return now;
  }
  int cnt = 0;
  auto it = now.begin();
  while (it != now.end() && cnt < c[cur]) {
    it++;
    cnt++;
  }
  if (cnt != c[cur]) {
    imposs = true;
    return now;
  }
  if (it == now.end()) {
    now.insert(it, {start++, cur});
    return now;
  }
  start++;
  now.insert(it, {it->first, cur});
  while (it != now.end()) {
    it->first++;
    it++;
  }
  return now;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int root = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x >> c[i];
    if (!x)
      root = i;
    else {
      adj[x].push_back(i);
      adj[i].push_back(x);
    }
  }
  memset(vis, 0, sizeof(vis));
  list<pair<int, int> > now = dfs(root);
  if (imposs) {
    cout << "NO";
    return 0;
  }
  int ans[n + 1];
  for (auto it = now.begin(); it != now.end(); it++) {
    ans[it->second] = it->first;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << ' ';
  }
}
