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;
double _eps = 1e-6;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
vector<int> adj[2000 + 100];
int arr[2000 + 100];
int f = 1;
void dfs(int root, vector<int> &v) {
  vector<int> nv;
  for (int x : adj[root]) {
    dfs(x, nv);
  }
  if (arr[root] <= nv.size()) {
    nv.insert(nv.begin() + arr[root], root);
  } else
    f = 0;
  for (int x : nv) v.push_back(x);
}
int main() {
  int n;
  cin >> n;
  int root;
  for (int i = 1; i <= n; i++) {
    int a, b;
    cin >> a >> arr[i];
    if (a == 0)
      root = i;
    else {
      adj[a].push_back(i);
    }
  }
  vector<int> v;
  dfs(root, v);
  if (!f)
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (int i = 1, k = 0; i <= n; ++i, k++) {
      arr[v[k]] = i;
    }
    for (int i = 1; i <= n; ++i) {
      cout << arr[i] << " ";
    }
    puts("");
  }
  return 0;
}
