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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int p[2005], c[2005], val[2005], t[2005];
vector<int> ch[2005];
vector<int> ord[2005];
void dfs(int v) {
  for (auto& x : ch[v]) {
    dfs(x);
    for (auto& y : ord[x]) {
      ord[v].push_back(y);
    }
  }
  if (ord[v].size() < c[v]) {
    cout << "NO";
    exit(0);
  }
  ord[v].insert(ord[v].begin() + c[v], v);
}
int ans[2005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i] >> c[i];
    ch[p[i]].push_back(i);
  }
  dfs(0);
  for (int i = 1; i <= n; i++) ans[ord[0][i]] = i;
  cout << "YES\n";
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << ' ';
  }
}
