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 double PI = 2 * acos(0.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());
const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";
template <class T0, class T1>
inline ostream &operator<<(ostream &out, pair<T0, T1> &a) {
  return out << "{" << a.first << ", " << a.second << "}";
}
template <class T0, class T1, class T2>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << "}";
}
template <class T0, class T1, class T2, class T3>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2, T3> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << ", " << get<3>(a) << "}";
}
template <class T>
inline ostream &operator<<(ostream &out, vector<T> &a) {
  out << "[";
  for (int i = 0; i < int(a.size()); ++i)
    out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
  return out;
}
void smain();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(12) << fixed;
  smain();
  return 0;
}
const int N = 5e3 + 100;
int p[N];
int c[N];
int a[N];
int parent[N];
vector<int> g[N];
int n;
void dfs(int v) {
  int ch = -1;
  for (int i = 0; i < N; ++i) {
    if (i - p[i] == c[v]) {
      ch = i;
      break;
    }
  }
  if (ch == -1) {
    cout << "NO";
    exit(0);
  }
  a[v] = ch;
  for (int i = ch; i < N; ++i) {
    p[i]++;
  }
  for (int to : g[v]) {
    dfs(to);
  }
}
void kek_dfs(int v, int bn, int &cnt) {
  cnt += a[v] < bn;
  for (int to : g[v]) {
    kek_dfs(to, bn, cnt);
  }
}
bool check() {
  for (int i = 1; i <= n; ++i) {
    int cnt = 0;
    kek_dfs(i, a[i], cnt);
    if (cnt != c[i]) return false;
  }
  return true;
}
void smain() {
  cin >> n;
  int root = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> parent[i] >> c[i];
    if (!parent[i]) {
      root = i;
    } else {
      g[parent[i]].push_back(i);
    }
  }
  dfs(root);
  if (!check()) {
    cout << "NO";
    return;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) {
    cout << a[i] + 1 << " ";
  }
}
