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;
using ll = long long;
const int inf = 1e9 + 7;
const ll longinf = 1LL << 60;
const ll mod = 1e9 + 7;
vector<int> v[2020];
int ans[2020], a[2020];
vector<pair<int, int>> dfs(int x, int& st) {
  vector<pair<int, int>> ret = {};
  for (auto to : v[x]) {
    auto e = dfs(to, st);
    ret.insert(ret.end(), e.begin(), e.end());
  }
  if (a[x] > ret.size()) {
    cout << "NO" << endl;
    exit(0);
  }
  vector<pair<int, int>> nr(1 + ret.size());
  for (int i = (int)(0); i < (int)(ret.size()); ++i) {
    if (i < a[x])
      nr[i] = ret[i];
    else if (i == a[x]) {
      nr[i] = {ret[i].first, x};
      ans[x] = ret[i].first;
      for (int j = (int)(i); j < (int)(ret.size()); ++j) {
        ans[ret[j].second]++;
        nr[j + 1] = {ans[ret[j].second], ret[j].second};
      }
      break;
    }
  }
  if (a[x] == ret.size()) {
    int s = nr.size();
    nr[s - 1] = {st, x};
    ans[x] = st;
  }
  st += 1;
  return nr;
}
int main() {
  int n;
  cin >> n;
  int p = -1;
  for (int i = (int)(0); i < (int)(n); ++i) {
    int x;
    cin >> x;
    if (x) {
      v[--x].push_back(i);
    } else
      p = i;
    cin >> a[i];
  }
  int st = 1;
  dfs(p, st);
  cout << "YES" << endl;
  for (int i = (int)(0); i < (int)(n); ++i) cout << ans[i] << endl;
  return 0;
}
