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 = 100009;
vector<int> vec[2009], seq[2009];
int c[2009], flag = true;
void dfs(int node) {
  int j, i, v, sz = vec[node].size(), id = 0;
  for (i = 0; i < sz && flag; i++) {
    v = vec[node][i];
    dfs(v);
    int szv = seq[v].size();
    for (j = 0; j < szv; j++) {
      if (id == c[node]) seq[node].push_back(node), id++;
      seq[node].push_back(seq[v][j]);
      id++;
    }
  }
  if (id == c[node])
    seq[node].push_back(node);
  else if (id < c[node])
    flag = false;
}
int main() {
  int n, i, p, root, ar[2009];
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &p, &c[i]);
    if (p == 0) {
      root = i;
      continue;
    }
    vec[p].push_back(i);
  }
  dfs(root);
  if (flag) {
    printf("YES\n");
    int sz = seq[root].size();
    for (i = 0; i < sz; i++) {
      ar[seq[root][i]] = i + 1;
    }
    for (i = 1; i <= n; i++) {
      printf("%d ", ar[i]);
    }
    printf("\n");
  } else
    printf("NO\n");
  return 0;
}
