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;
int n, root;
vector<vector<int>> g;
vector<int> c, sz, res, used;
void calc_sz(int v) {
  for (auto ch : g[v]) {
    calc_sz(ch);
    sz[v] += sz[ch];
  }
  ++sz[v];
}
void getans(int v) {
  if (c[v] > sz[v] - 1) {
    cout << "NO" << endl;
    exit(0);
  }
  vector<int> vals;
  vals.reserve(sz[v]);
  for (int i = 1; i <= n; ++i) {
    if (!used[i] && vals.size() != sz[v]) {
      vals.push_back(i);
    }
  }
  res[v] = vals[c[v]];
  used[res[v]] = 1;
  for (auto ch : g[v]) {
    getans(ch);
  }
}
int main() {
  cin >> n;
  g.resize(n);
  c.resize(n);
  sz.resize(n);
  res.resize(n);
  used.resize(n + 1);
  for (int i = 0; i < n; ++i) {
    int pr;
    cin >> pr >> c[i];
    if (pr) {
      g[pr - 1].push_back(i);
    } else {
      root = i;
    }
  }
  calc_sz(root);
  getans(root);
  cout << "YES" << endl;
  for (auto x : res) {
    cout << x << " ";
  }
}
