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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
vector<vector<long long int> > grid;
vector<long long int> cnt, ans;
bool possible = true;
vector<pair<long long int, long long int> > dfs(long long int u,
                                                long long int v) {
  vector<pair<long long int, long long int> > temp;
  if (!possible) return temp;
  for (auto &x : grid[u]) {
    if (x == v) continue;
    if (!possible) return temp;
    vector<pair<long long int, long long int> > t = dfs(x, u);
    for (auto &x : t) temp.push_back(x);
  }
  if (temp.size() < cnt[u]) {
    possible = false;
    return temp;
  }
  sort(temp.begin(), temp.end());
  pair<long long int, long long int> t = {cnt[u] + 1, u};
  temp.insert(temp.begin() + cnt[u], t);
  for (long long int i = 0; i < temp.size(); i++) {
    temp[i].first = i + 1;
  }
  return temp;
}
void solve() {
  long long int n;
  cin >> n;
  grid.resize(n + 1);
  cnt.resize(n + 1);
  long long int r;
  for (long long int i = 1; i < n + 1; i++) {
    long long int a;
    cin >> a >> cnt[i];
    if (a == 0) {
      r = i;
      continue;
    }
    grid[i].push_back(a);
    grid[a].push_back(i);
  }
  vector<pair<long long int, long long int> > temp = dfs(r, r);
  if (!possible) {
    cout << "NO"
         << "\n";
    return;
  }
  cout << "YES"
       << "\n";
  ans.resize(n + 1);
  for (auto &x : temp) ans[x.second] = x.first;
  for (long long int i = 1; i < n + 1; i++) cout << ans[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  for (long long int i = 0; i < t; i++) {
    solve();
  }
}
