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;
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  out << '(' << p.first << ',' << p.second << ')';
  return out;
}
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
template <class T>
istream& operator>>(istream& in, vector<T>& v) {
  for (auto& x : v) in >> x;
  return in;
}
template <class T>
ostream& operator<<(ostream& out, vector<vector<T>>& v) {
  for (auto& x : v) out << x << '\n';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, vector<T>& v) {
  for (auto x : v) out << x << ' ';
  return out;
}
long long gcd(long long a, long long b) {
  if (b > a) swap(a, b);
  return (b ? gcd(b, a % b) : a);
}
using ll = long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using tiii = pair<pair<int, int>, int>;
using vi = vector<int>;
using vl = vector<long long>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<long long>>;
const int h = 1000000007;
vi ins, l;
vvi M;
vi C, ans;
void dfs1(int x) {
  l[x] = 1;
  for (int y : M[x]) dfs1(y), l[x] += l[y];
}
void dfs(int x) {
  if (C[x] >= l[x]) {
    cout << "NO";
    exit(0);
  }
  ans[x] = ins[C[x]];
  ins.erase(ins.begin() + C[x]);
  for (int y : M[x]) dfs(y);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cerr.setstate(ios::failbit);
  int n;
  cin >> n;
  ans.resize(n + 1);
  M.resize(n + 1);
  ins.resize(n);
  C.resize(n + 1);
  l.resize(n + 1);
  for (int i = 0; i < n; i++) ins[i] = i + 1;
  for (int i = 1; i < n + 1; i++) {
    int p, c;
    cin >> p >> c;
    M[p].push_back(i);
    C[i] = c;
  }
  dfs1(M[0][0]);
  dfs(M[0][0]);
  cout << "YES\n";
  for (int i = 1; i < n + 1; i++) cout << ans[i] << ' ';
}
