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;
vector<int> a[2001];
double ans[2001];
int c[2001];
int un;
set<double> f(int root, int p) {
  set<double> s;
  for (int e : a[root]) {
    if (e != p) {
      for (double val : f(e, root)) s.insert(val);
    }
  }
  if (s.size() < c[root]) {
    cout << "NO";
    exit(0);
  }
  if (s.empty()) {
    ans[root] = un;
    un += 200000;
  } else if (c[root] == 0) {
    ans[root] = *s.begin() - 10000;
  } else {
    auto it = s.begin();
    while (c[root]--) ++it;
    auto itt = it;
    --itt;
    if (it == s.end())
      ans[root] = *itt + 10000;
    else
      ans[root] = (*itt + *it) / 2;
  }
  s.insert(ans[root]);
  return s;
}
int main() {
  cout << fixed << setprecision(10);
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  un = 100000;
  cin >> n;
  int root;
  for (int i = 1; i < n + 1; i++) {
    int p;
    cin >> p >> c[i];
    if (p == 0)
      root = i;
    else
      a[p].push_back(i);
  }
  f(root, -1);
  pair<double, int> b[n];
  for (int i = 1; i < n + 1; i++) b[i - 1] = make_pair(ans[i], i);
  sort(b, b + n);
  int cur = 1;
  for (auto e : b) ans[e.second] = cur++;
  cout << "YES\n";
  for (int i = 1; i < n + 1; i++) cout << (int)ans[i] << ' ';
}
