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 long NN = 2e3 + 5;
long long n, i, p[NN], a[NN], jaw[NN], R, ketemu;
vector<long long> v[NN];
void dfs(long long pos) {
  if (ketemu) return;
  if (a[pos] == 0) {
    ketemu = pos;
    return;
  }
  for (long long i = 0; i < v[pos].size(); i++) dfs(v[pos][i]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long ta, tb;
    cin >> p[i] >> a[i];
    if (p[i] == 0)
      R = i;
    else
      v[p[i]].push_back(i);
  }
  for (long long i = 1; i <= n; i++) {
    ketemu = 0;
    dfs(R);
    if (ketemu == 0) {
      cout << "NO\n";
      return 0;
    }
    jaw[ketemu] = i;
    while (ketemu) {
      a[ketemu]--;
      ketemu = p[ketemu];
    }
  }
  cout << "YES\n";
  for (long long i = 1; i <= n; i++) cout << jaw[i] << " ";
  cout << "\n";
}
