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>
#pragma GCC optimize("O3")
using namespace std;
const int N = (int)2e5 + 100;
const int inf = 1e9 + 100;
const double eps = 1e-9;
vector<int> v[N];
int c[N];
int ans[N];
vector<int> dfs(int x) {
  vector<int> t;
  for (auto y : v[x]) {
    vector<int> c = dfs(y);
    for (auto j : c) {
      t.push_back(j);
    }
  }
  if (t.size() < c[x]) {
    cout << "NO";
    exit(0);
  }
  t.push_back(0);
  int i;
  for ((i) = t.size() - 1; (i) >= 0; (i)--) {
    if (i == c[x]) {
      t[i] = x;
      break;
    }
    t[i] = t[i - 1];
  }
  return t;
}
int main() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  srand(time(0));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, i, p;
  cin >> n;
  int kor = 0;
  for ((i) = 0; (i) < (n); (i)++) {
    cin >> p >> c[i];
    if (!p) {
      kor = i;
    } else {
      v[p - 1].push_back(i);
    }
  }
  vector<int> c = dfs(kor);
  cout << "YES\n";
  for ((i) = 0; (i) < (c.size()); (i)++) {
    ans[c[i]] = i + 1;
  }
  for ((i) = 0; (i) < (n); (i)++) {
    cout << ans[i] << ' ';
  }
}
