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 int N = 1e5;
long long arr[N], P[N], C[N];
vector<int> adj[N];
vector<pair<int, int> > res[N];
int sol[N];
bool valid = true;
void dfs(int v) {
  vector<pair<int, int> > temp;
  for (int u : adj[v]) {
    dfs(u);
    for (pair<int, int> s : res[u]) res[v].push_back(s);
  }
  sort(res[v].begin(), res[v].end());
  if (C[v] > res[v].size()) {
    valid = false;
  } else {
    long long val = 0;
    if (C[v] == 0) {
      val = 1;
    } else {
      val = res[v][C[v] - 1].first + 1;
    }
    for (int i = C[v]; i < res[v].size(); ++i) {
      res[v][i].first += 2;
    }
    res[v].push_back(make_pair(val, v));
    sort(res[v].begin(), res[v].end());
  }
}
int main() {
  int n;
  cin >> n;
  int root = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> P[i] >> C[i];
    adj[P[i]].push_back(i);
    if (P[i] == 0) root = i;
  }
  dfs(root);
  for (pair<int, int> u : res[root]) {
    sol[u.second] = u.first;
  }
  if (!valid) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; ++i) cout << sol[i] << " ";
  cout << endl;
}
