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;
vector<int> ans[2020];
int N, R;
int C[2020];
vector<int> adj[2020];
int sz[2020];
int pos[2020];
void dfs(int v, int p) {
  sz[v] = 1;
  for (int i : adj[v]) {
    if (i == p) continue;
    dfs(i, v);
    sz[v] += sz[i];
  }
  if (C[v] > sz[v] - 1) {
    puts("NO");
    exit(0);
  }
  for (int i : adj[v]) {
    if (i == p) continue;
    for (int j : ans[i]) {
      if (ans[v].size() == C[v]) ans[v].push_back(v);
      ans[v].push_back(j);
    }
  }
  if (ans[v].size() == C[v]) ans[v].push_back(v);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    int p;
    scanf("%d%d", &p, &C[i]);
    if (p) {
      adj[p].push_back(i);
      adj[i].push_back(p);
    } else
      R = i;
  }
  dfs(R, 0);
  puts("YES");
  for (int i = 0; i < ans[R].size(); i++) pos[ans[R][i]] = i + 1;
  for (int i = 1; i <= N; i++) printf("%d ", pos[i]);
  puts("");
  return 0;
}
