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>
const double PI = 3.141592653589793238460;
using namespace std;
vector<int> ar[2001], col;
int c[2001], vis[2001], root, sub[2001], res[2001];
int getKth(int k) {
  int res = col[k];
  for (int i = k; i + 1 < col.size(); i++) col[i] = col[i + 1];
  col.pop_back();
  return res;
}
bool dfs(int node, int par = -1) {
  if (sub[node] - 1 < c[node]) return false;
  res[node] = getKth(c[node]);
  for (int child : ar[node])
    if (child != par) {
      if (dfs(child, node) == false) return false;
    }
  return true;
}
int dfs1(int node, int par) {
  sub[node] = 1;
  for (int child : ar[node])
    if (child != par) sub[node] += dfs1(child, node);
  return sub[node];
}
int main() {
  int n, p, x;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    col.push_back(i);
    cin >> p >> x;
    if (p == 0)
      root = i, c[i] = x;
    else
      c[i] = x, ar[i].push_back(p), ar[p].push_back(i);
  }
  dfs1(root, -1);
  bool ans = dfs(root);
  if (ans == false)
    cout << "NO";
  else {
    cout << "YES\n";
    for (int i = 1; i <= n; i++) cout << res[i] << " ";
  }
}
