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 maxn = 1e5 + 10;
set<pair<int, int>> adads;
int c[maxn];
int mg[maxn];
vector<int> roots;
int n;
vector<int> adj[maxn];
void in();
void solve();
set<pair<int, int>> dfs(int u);
void out();
int main() {
  in();
  solve();
  out();
}
void in() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p >> c[i];
    p--;
    if (p == -1)
      roots.push_back(i);
    else
      adj[p].push_back(i);
  }
}
void solve() {
  for (auto i : roots) {
    adads = dfs(i);
    for (auto it = adads.begin(); it != adads.end(); it++) {
      auto x = *it;
      mg[x.second] = x.first;
    }
  }
}
void out() {
  cout << "YES\n";
  for (int i = 0; i < n; i++) cout << mg[i] + 1 << ' ';
}
set<pair<int, int>> dfs(int u) {
  set<pair<int, int>> adads;
  for (auto v : adj[u]) {
    auto wow = dfs(v);
    for (auto it = wow.begin(); it != wow.end(); it++) adads.insert(*it);
  }
  if (adads.size() < c[u]) {
    cout << "NO";
    exit(0);
  }
  if (adads.size() == c[u]) {
    if (adads.size() == 0) {
      adads.insert({0, u});
      return adads;
    }
    auto it = adads.end();
    it--;
    adads.insert({(*it).first + 1, u});
    return adads;
  }
  auto it = adads.begin();
  set<pair<int, int>> tmp;
  for (int i = 0; i < c[u]; i++) {
    tmp.insert(*it);
    it++;
  }
  mg[u] = (*it).first + 1;
  for (int i = 0; it != adads.end(); i++) {
    auto x = *it;
    tmp.insert({x.first + 2, x.second});
    it++;
  }
  adads = tmp;
  adads.insert({mg[u], u});
  return adads;
}
