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 + 100;
int c[N] = {0}, vis[N] = {0}, ans[N] = {0}, sv[N] = {0};
set<int> s;
vector<int> g[N];
int n;
void dfscheck(int node) {
  sv[node] = 1;
  for (auto it : g[node]) {
    dfscheck(it);
    sv[node] += sv[it];
  }
}
void dfs(int node) {
  if (sv[node] <= c[node]) {
    cout << "NO\n";
    exit(0);
  }
  auto it = s.begin();
  for (int i = 0; i < c[node]; i++) it++;
  ans[node] = *it;
  s.erase(it);
  for (auto it2 : g[node]) dfs(it2);
}
int main() {
  int root = 0, i, x;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x >> c[i];
    if (x == 0) root = i;
    g[x].push_back(i);
  }
  for (i = 1; i <= n; i++) s.insert(i);
  dfscheck(root);
  dfs(root);
  cout << "YES\n";
  for (i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}
