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>
#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
vector<long long> v[2005];
long long ar[2005];
bool ch = true;
vector<long long> dfs(long long x) {
  vector<long long> te;
  for (long long i = 0; i < v[x].size(); i++) {
    vector<long long> rr = dfs(v[x][i]);
    for (long long j = 0; j < rr.size(); j++) {
      te.push_back(rr[j]);
    }
  }
  if (ar[x] > te.size()) {
    ch = false;
    cout << "NO";
    exit(0);
  }
  te.insert(te.begin() + ar[x], x);
  return te;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, pa;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long x, y;
    cin >> x >> y;
    if (x == 0)
      pa = i;
    else
      v[x].push_back(i);
    ar[i] = y;
  }
  vector<long long> v1 = dfs(pa);
  long long re[n + 1];
  long long cc = 1;
  if (!ch) {
    cout << "NO";
    return 0;
  }
  for (long long i = 0; i < v1.size(); i++) {
    re[v1[i]] = cc;
    cc++;
  }
  cout << "YES" << endl;
  for (long long i = 1; i <= n; i++) {
    cout << re[i] << " ";
  }
}
