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;
vector<vector<int>> sons;
vector<long double> v;
vector<int> c;
long double t;
bool valid;
set<long double> solve(int id) {
  set<long double> ss = {0.0, 10000007.0 * 2050.0 * 2050.0}, s;
  if (sons[id].empty()) {
    if (c[id] != 0) valid = false;
    v[id] = t;
    ss.insert(t);
    t += 10000007.0 * (rand() % 10000) * 0.0001;
    return ss;
  } else {
    for (auto p : sons[id]) {
      s = solve(p);
      for (auto pp : s) ss.insert(pp);
    }
    if (c[id] > ss.size() - 2) {
      valid = false;
      return ss;
    } else {
      long double x, y;
      int cc = 0;
      for (auto p : ss) {
        if (cc == c[id]) x = p;
        if (cc == c[id] + 1) y = p;
        cc++;
      }
      v[id] = (x + y) * .5;
      ss.insert(v[id]);
    }
    return ss;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, i, j, k, p;
  cin >> n;
  c.resize(n + 1);
  sons.resize(n + 1);
  v.resize(n + 1);
  for (i = 1; i <= n; i++) {
    cin >> p >> c[i];
    sons[p].push_back(i);
  }
  t = 10000007.0;
  valid = true;
  set<long double> s = solve(sons[0][0]);
  if (!valid)
    cout << "NO\n";
  else {
    cout << "YES\n";
    vector<int> ans(n + 1);
    vector<pair<long double, int>> vv;
    for (i = 1; i <= n; i++) vv.push_back({v[i], i});
    sort(vv.begin(), vv.end());
    for (i = 0; i < n; i++) ans[vv[i].second] = i + 1;
    for (i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << "\n";
  }
  return 0;
}
