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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int N = 2005;
int n, s[N], c[N], a[N];
vector<int> e[N];
int sf(int first) {
  s[first] = 1;
  for (int i : e[first]) s[first] += sf(i);
  return s[first];
}
void f(int first, vector<int> v) {
  if (v.size() <= c[first]) {
    cout << "NO";
    exit(0);
  }
  a[first] = v[c[first]];
  int j = 0;
  for (int i : e[first]) {
    vector<int> w;
    for (int k = 0; k < s[i]; k++) {
      if (j == c[first]) j++;
      w.push_back(v[j++]);
    }
    f(i, w);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  int root;
  for (int i = 1; i <= n; i++) {
    int par;
    cin >> par >> c[i];
    e[par].push_back(i);
    if (!par) root = i;
  }
  sf(root);
  vector<int> v;
  for (int i = 1; i <= n; i++) v.push_back(i);
  f(root, v);
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << a[i] << ' ';
  cout << '\n';
  return 0;
}
