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 = 2e3 + 10;
vector<int> E[N];
int a[N], c[N];
int n;
vector<int> dfs(int x) {
  vector<int> ans;
  for (int i = 0; i < E[x].size(); i++) {
    vector<int> tmp = dfs(E[x][i]);
    for (int j = 0; j < tmp.size(); j++) {
      ans.push_back(tmp[j]);
    }
  }
  if (x == 0) return ans;
  if (ans.size() < c[x]) {
    puts("NO");
    exit(0);
  }
  vector<int> ans_;
  for (int i = 0; i < c[x]; i++) {
    ans_.push_back(ans[i]);
  }
  ans_.push_back(x);
  for (int i = c[x]; i < ans.size(); i++) {
    ans_.push_back(ans[i]);
  }
  return ans_;
}
int main() {
  scanf("%d", &n);
  int fa;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &fa, c + i);
    E[fa].push_back(i);
  }
  vector<int> ans = dfs(0);
  puts("YES");
  for (int i = 0; i < ans.size(); i++) {
    a[ans[i]] = i + 1;
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", a[i]);
  }
  puts("");
  return 0;
}
