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 int M = 1e9 + 7;
const long long int mod = M;
const long long int N = 4e3 + 7;
long long int a[N], b[N], c[N];
vector<long long int> adj[N];
vector<pair<long long int, long long int> > v[N], ans;
long long int vist[N], root;
long long int val = 0, n;
void dfs(long long int src) {
  for (auto x : adj[src]) {
    dfs(x);
    for (auto y : v[x]) v[src].push_back(y);
  }
  val++;
  if (adj[src].size() == 0) {
    if (b[src] != 0) {
      cout << "NO";
      exit(0);
    } else {
      v[src].push_back({val, src});
      c[src] = val;
    }
    return;
  }
  sort(v[src].begin(), v[src].end());
  if (v[src].size() < b[src]) {
    cout << "NO";
    exit(0);
  }
  long long int ins;
  if (b[src] == 0)
    ins = v[src][b[src]].first;
  else
    ins = v[src][b[src] - 1].first + 1;
  for (long long int i = b[src]; i < v[src].size(); i++) {
    v[src][i].first += 1;
  }
  v[src].push_back({ins, src});
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    if (a[i] != 0) {
      adj[a[i]].push_back(i);
    } else {
      root = i;
    }
  }
  dfs(root);
  for (auto x : v[root]) {
    ans.push_back({x.second, x.first});
  }
  sort(ans.begin(), ans.end());
  cout << "YES\n";
  for (auto x : ans) {
    cout << x.second << " ";
  }
}
