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;
#pragma GCC optimize("-O3")
const int MAXN = (int)3e3;
vector<int> G[MAXN];
vector<int> kolejnosc[MAXN];
int C[MAXN];
int odp[MAXN];
int n;
int root;
void concat(vector<int> &A, vector<int> B) {
  for (auto b : B) A.push_back(b);
}
void koniec() {
  cout << "NO"
       << "\n";
  exit(0);
}
void dfs(int x) {
  for (int y : G[x]) {
    dfs(y);
    concat(kolejnosc[x], kolejnosc[y]);
  }
  if (kolejnosc[x].size() < C[x]) koniec();
  kolejnosc[x].insert(kolejnosc[x].begin() + C[x], x);
}
void input() {
  cin >> n;
  for (int i = 0; i < (n); ++i) {
    int a;
    cin >> a;
    if (a > 0) {
      a--;
      G[a].push_back(i);
    } else
      root = i;
    cin >> C[i];
  }
}
int main() {
  ios_base::sync_with_stdio(0), cout.setf(ios::fixed), cout.precision(10);
  input();
  dfs(root);
  cout << "YES"
       << "\n";
  for (int i = 0; i < (n); ++i) {
    odp[kolejnosc[root][i]] = i + 2;
  }
  for (int i = 0; i < (n); ++i) cout << odp[i] << " ";
  cout << "\n";
}
