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;
void die() {
  cout << "NO\n";
  exit(0);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> G(n + 1);
  int root = -1;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p >> a[i];
    if (p == 0)
      root = i;
    else
      G[p].push_back(i), G[i].push_back(p);
  }
  if (root == -1) die();
  vector<int> st_size(n + 1);
  vector<int> ans(n + 1, -1);
  vector<int> used(n + 1, 0);
  function<void(int, int)> dfs = [&](int u, int p) -> void {
    int cnt = 0, last = 0;
    for (int x = 1; x <= n; x++) {
      if (used[x]) continue;
      cnt++;
      if (cnt == a[u] + 1) {
        last = x;
        break;
      }
    }
    ans[u] = last;
    used[last] = 1;
    st_size[u] = 1;
    for (int v : G[u]) {
      if (v == p) continue;
      dfs(v, u);
      st_size[u] += st_size[v];
    }
    if (!(a[u] <= st_size[u] - 1)) die();
  };
  dfs(root, -1);
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << ans[i] << " \n"[i == n];
  return (0);
}
