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;
int n;
vector<int> c, a;
vector<vector<int> > g;
vector<int> dfs(int u) {
  vector<int> subtree;
  for (auto& v : g[u]) {
    vector<int> tmp = dfs(v);
    for (auto& x : tmp) {
      subtree.push_back(x);
    }
  }
  if (c[u] > (int)subtree.size()) {
    cout << "NO\n";
    exit(0);
  } else {
    sort(subtree.begin(), subtree.end(),
         [&](const int& lhs, const int& rhs) { return a[lhs] < a[rhs]; });
    if (g[u].empty()) {
      a[u] = 1;
    } else {
      if (c[u] == 0) {
        a[u] = 1;
      } else {
        a[u] = a[subtree[c[u] - 1]] + 1;
      }
      for (int j = c[u]; j < (int)subtree.size(); j++) {
        a[subtree[j]] += 5;
      }
    }
  }
  subtree.push_back(u);
  return subtree;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  c.resize(n + 1);
  g.resize(n + 1);
  int root;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p >> c[i];
    if (p == 0) {
      root = i;
    } else {
      g[p].push_back(i);
    }
  }
  a.resize(n + 1);
  dfs(root);
  cout << "YES\n";
  for (int i = 1; i <= n; i++) {
    cout << a[i] << ' ';
  }
  cout << '\n';
  return 0;
}
