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 long double PI = acos(-1.0);
template <class T>
using min_queue = priority_queue<T, vector<T>, greater<T>>;
template <class T>
int sz(const T &first) {
  return (int)first.size();
}
template <class T>
ostream &operator<<(ostream &os, vector<T> &v) {
  os << "\n[";
  for (T &first : v) os << first << ',';
  return os << "]\n";
}
struct pairhash {
 public:
  template <typename T1, typename T2>
  size_t operator()(const pair<T1, T2> &p) const {
    size_t lhs = hash<T1>()(p.first);
    size_t rhs = hash<T2>()(p.second);
    return lhs ^ (rhs + 0x9e3779b9 + (lhs << 6) + (lhs >> 2));
  }
};
bool pos = true;
vector<long long> A, rem;
vector<long long> C;
vector<long long> S;
vector<vector<long long>> adj;
void dfs1(long long i) {
  S[i] = 1;
  for (long long c : adj[i]) {
    dfs1(c);
    S[i] += S[c];
  }
}
void dfs2(long long i) {
  if (C[i] >= S[i]) {
    pos = false;
    return;
  }
  A[i] = rem[C[i]];
  rem.erase(rem.begin() + C[i]);
  for (long long c : adj[i]) {
    dfs2(c);
  }
}
void run() {
  long long n;
  cin >> n;
  A = vector<long long>(n, 0);
  C = vector<long long>(n, 0);
  S = vector<long long>(n, 0);
  adj = vector<vector<long long>>(n);
  for (long long i = 1; i <= n; i++) {
    rem.push_back(i);
  }
  long long root = 0;
  for (long long i = 0; i < n; i++) {
    long long p, c;
    cin >> p >> c;
    C[i] = c;
    if (p == 0) {
      root = i;
    } else {
      adj[p - 1].push_back(i);
    }
  }
  dfs1(root);
  dfs2(root);
  if (pos) {
    cout << "YES" << endl;
    for (long long i = 0; i < n; i++) {
      cout << A[i] << " ";
    }
    cout << endl;
  } else {
    cout << "NO" << endl;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(20);
  run();
  return 0;
}
