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 int N = 2100;
int n, root, c[N], a[N], cnt[N];
vector<int> adj[N];
void Dfs1(int u, int p) {
  cnt[u] = 1;
  for (int v : adj[u])
    if (v != p) {
      Dfs1(v, u);
      cnt[u] += cnt[v];
    }
}
void Dfs2(int u, int p, vector<int> num) {
  if (c[u] >= cnt[u]) {
    cout << "NO";
    exit(0);
  }
  a[u] = num[c[u]];
  num[c[u]] = 0;
  for (int v : adj[u])
    if (v != p) {
      vector<int> tmp;
      for (int i = 1; i <= cnt[v]; i++) {
        if (num.back() == 0) {
          num.pop_back();
          i--;
        } else {
          tmp.push_back(num.back());
          num.pop_back();
        }
      }
      reverse(tmp.begin(), tmp.end());
      Dfs2(v, u, tmp);
    }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p >> c[i];
    if (p == 0)
      root = i;
    else {
      adj[i].push_back(p);
      adj[p].push_back(i);
    }
  }
  vector<int> tmp;
  for (int i = 1; i <= n; i++) tmp.push_back(i);
  Dfs1(root, root);
  Dfs2(root, root, tmp);
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) cout << a[i] << ' ';
}
