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 = 2005;
vector<int> operator+(vector<int> a, vector<int> b) {
  for (auto x : b) a.push_back(x);
  return a;
}
int n, rt;
int c[N];
vector<int> G[N];
vector<int> dfs(int p) {
  vector<int> rt;
  for (auto e : G[p]) rt = rt + dfs(e);
  if (!(c[p] <= (signed)rt.size())) {
    puts("NO");
    exit(0);
  }
  rt.insert(rt.begin() + c[p], p);
  return rt;
}
int main() {
  scanf("%d", &n);
  for (int i = 1, fa; i <= n; i++) {
    scanf("%d%d", &fa, c + i);
    if (fa) {
      G[fa].push_back(i);
    } else {
      rt = i;
    }
  }
  vector<int> p = dfs(rt);
  for (int i = 0; i < n; i++) c[p[i]] = i + 1;
  puts("YES");
  for (int i = 1; i <= n; i++) printf("%d ", c[i]);
  return 0;
}
