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;
double pi = 3.141592653589793238462;
string numkey = "0123456789";
string uni = "abcdefghijklmnopqrstuvwxyz";
long long int dx[4] = {0, 0, 1, -1};
long long int dy[4] = {1, -1, 0, 0};
long long int INF = (1LL << 62);
long long int c[10000], a[10000];
vector<long long int> g[10000];
vector<long long int> dfs(long long int u) {
  vector<long long int> u_sons;
  for (auto v : g[u]) {
    vector<long long int> v_sons = dfs(v);
    u_sons.insert(u_sons.end(), v_sons.begin(), v_sons.end());
  }
  if (c[u] > u_sons.size()) {
    cout << "NO\n";
    exit(0);
  }
  u_sons.insert(u_sons.begin() + c[u], u);
  return u_sons;
}
signed main() {
  long long int n;
  cin >> n;
  for (long long int i = 1; i <= n; i++) {
    long long int p;
    cin >> p >> c[i];
    g[p].push_back(i);
  }
  vector<long long int> v = dfs(0);
  for (long long int i = 1; i <= n; i++) a[v[i]] = i;
  cout << "YES" << endl;
  for (long long int i = 1; i <= n; i++) {
    cout << a[i] << " ";
  }
  return 0;
}
