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 long long MX = 2020;
long long n, p, c[MX], rt, vis[MX], chk, ans[MX];
vector<long long> v[MX];
vector<long long> ord[MX];
void dfs(long long x) {
  vis[x] = 1;
  for (int i = 0; i < v[x].size(); i++) {
    long long next = v[x][i];
    if (!vis[next]) {
      dfs(next);
      for (int j = 0; j < ord[next].size(); j++) {
        ord[x].push_back(ord[next][j]);
      }
    }
  }
  if (ord[x].size() < c[x]) {
    chk = 1;
    return;
  }
  ord[x].insert(ord[x].begin() + c[x], x);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p >> c[i];
    if (p != 0) {
      v[p].push_back(i);
      v[i].push_back(p);
    } else
      rt = i;
  }
  dfs(rt);
  if (chk)
    cout << "NO\n";
  else {
    for (int i = 0; i < ord[rt].size(); i++) {
      ans[ord[rt][i]] = i + 1;
    }
    cout << "YES\n";
    for (int i = 1; i <= n; i++) {
      cout << ans[i] << " ";
    }
  }
}
