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 = 2005;
int c[N];
int a[N];
int sub[N];
vector<int> g[N];
void dfs(int u) {
  sub[u] = 1;
  for (int v : g[u]) {
    dfs(v);
    sub[u] += sub[v];
  }
  if (sub[u] <= c[u]) {
    cout << "NO";
    exit(0);
  }
}
void solve(int u, vector<int> values) {
  a[u] = values[c[u]];
  values.erase(values.begin() + c[u]);
  for (int v : g[u]) {
    vector<int> t;
    while (t.size() < sub[v]) {
      t.push_back(values.back());
      values.pop_back();
    }
    reverse(t.begin(), t.end());
    solve(v, t);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int root = -1;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p >> c[i];
    if (p == 0) root = i;
    if (p) g[p].push_back(i);
  }
  dfs(root);
  vector<int> v;
  for (int i = 1; i <= n; i++) v.push_back(i);
  solve(root, v);
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) {
    cout << a[i] << " ";
  }
  return 0;
}
